LeetCode中关于字符串的题目的整理（待更）
检测大写字母（simple难度）
https://leetcode-cn.com/problems/detect-capital/
本题方法和代码来源：
作者：linaxiaokeai163com
链接：https://leetcode-cn.com/problems/detect-capital/solution/jian-ce-da-xiao-zi-fu-zai-suo-you-java-t-9ibf/
来源：力扣（LeetCode）
class Solution {
public boolean detectCapitalUse(String word) {
boolean res = false;
int count = 0;
char x;
for (int i = 0; i < word.length(); i++) {
x = word.charAt(i);
if(x >= 65 && x <= 90) {
count++;
}
// if (word.charAt(i) >= 97 && word.charAt(i) <= 122) {
// countMin++;
// }
}
if ((count == 1 && (word.charAt(0) >= 65 && word.charAt(0) <= 90)) || (count == word.length())) {
res = true;
} else if (count == 0) {
res = true;
}
return res;
}
}
作者：linaxiaokeai163com
链接：https://leetcode-cn.com/problems/detect-capital/solution/jian-ce-da-xiao-zi-fu-zai-suo-you-java-t-9ibf/
来源：力扣（LeetCode）反转字符串（simple）
https://leetcode-cn.com/problems/reverse-string/
class Solution {
public void reverseString(char[] s) {
if(s.length == 0){
return;
}
int left = 0;
int right = s.length - 1;
while(left < right){
char temp = s[left];
s[left] = s[right];
s[right] = temp;
left++;
right--;
}
}
}反转字符串Ⅱ（simple难度）
https://leetcode-cn.com/problems/reverse-string-ii/
注意事项：char型数组转换成字符串时不要直接调用toString方法，用Stirng
类的构造器方法来实现。
代码逻辑见注释
class Solution {
public String reverseStr(String s, int k) {
char[] charArray = s.toCharArray();
k >= 字符串长度：直接整个字符串反转
if (k >= s.length()) {
return new String(swap(charArray, 0, s.length() - 1));
}
int step = 0;
int len;
//若下一个2k间隔>=字符串长度，则反转完成
while (step < s.length()) {
//len的取值判断防止调用swap方法时发生下标越界
len = step + k - 1 >= s.length() ? s.length() - 1 : step + k - 1;
反转[step,len]范围内的元素
charArray = swap(charArray, step, len);
//step：每次增加2*k
step += 2 * k;
}
return new String(charArray);
}
//对指定下标范围内的元素进行反转
private char[] swap(char[] charArray, int left, int right) {
while (left < right) {
char temp = charArray[left];
charArray[left] = charArray[right];
charArray[right] = temp;
left++;
right--;
}
return charArray;
}
}
作者：HIT_whc
链接：https://leetcode-cn.com/problems/reverse-string-ii/solution/javashi-jian-ji-bai-100-by-hit_whc-o8u9/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。翻转字符串里面的单词（medium难度）
https://leetcode-cn.com/problems/reverse-words-in-a-string/
class Solution {
public String reverseWords(String s) {
// 除去开头和末尾的空白字符
s = s.trim();
// 正则匹配连续的空白字符作为分隔符分割
List wordList = Arrays.asList(s.split("\\s+"));
Collections.reverse(wordList);
return String.join(" ", wordList);
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/reverse-words-in-a-string/solution/fan-zhuan-zi-fu-chuan-li-de-dan-ci-by-leetcode-sol/
来源：力扣（LeetCode） 本题方法和代码来源：
作者：jyd
链接：https://leetcode-cn.com/problems/reverse-words-in-a-string/solution/151-fan-zhuan-zi-fu-chuan-li-de-dan-ci-shuang-zh-2/
来源：力扣（LeetCode）
class Solution {
public String reverseWords(String s) {
s = s.trim(); // 删除首尾空格
int j = s.length() - 1, i = j;
StringBuilder res = new StringBuilder();
while(i >= 0) {
while(i >= 0 && s.charAt(i) != ' ') i--; // 搜索首个空格
res.append(s.substring(i + 1, j + 1) + " "); // 添加单词
while(i >= 0 && s.charAt(i) == ' ') i--; // 跳过单词间空格
j = i; // j 指向下个单词的尾字符
}
return res.toString().trim(); // 转化为字符串并返回
}
}
作者：jyd
链接：https://leetcode-cn.com/problems/reverse-words-in-a-string/solution/151-fan-zhuan-zi-fu-chuan-li-de-dan-ci-shuang-zh-2/
来源：力扣（LeetCode）
class Solution {
public String reverseWords(String s) {
String[] strs = s.trim().split(" "); // 删除首尾空格，分割字符串
StringBuilder res = new StringBuilder();
for(int i = strs.length - 1; i >= 0; i--) { // 倒序遍历单词列表
if(strs[i].equals("")) continue; // 遇到空单词则跳过
res.append(strs[i] + " "); // 将单词拼接至 StringBuilder
}
return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回
}
}
作者：jyd
链接：https://leetcode-cn.com/problems/reverse-words-in-a-string/solution/151-fan-zhuan-zi-fu-chuan-li-de-dan-ci-shuang-zh-2/
来源：力扣（LeetCode）反转字符串中的单词Ⅲ（simple难度）
https://leetcode-cn.com/problems/reverse-words-in-a-string-iii/
class Solution {
public String reverseWords(String s) {
char[] sArray = s.toCharArray();
int i = 0, j = 0;
while(j < sArray.length){
//单词开头
while(i < sArray.length && sArray[i] == ' '){
i++;
}
j = i;
//单词结尾
while(j < sArray.length && sArray[j] != ' '){
j++;
}
reverse(sArray, i, j-1);
i = j;
}
return String.valueOf(sArray);
}
public void reverse(char[] s, int start, int end){
while(start < end){
char temp = s[start];
s[start] = s[end];
s[end] = temp;
start++;
end--;
}
}
}验证回文串（simple难度）
https://leetcode-cn.com/problems/valid-palindrome/
在原字符串上直接判断
class Solution {
public boolean isPalindrome(String s) {
int n = s.length();
int left = 0, right = n - 1;
while (left < right) {
while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
++left;
}
while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
--right;
}
if (left < right) {
if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
return false;
}
++left;
--right;
}
}
return true;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/valid-palindrome/solution/yan-zheng-hui-wen-chuan-by-leetcode-solution/
来源：力扣（LeetCode）最长公共前缀（simple难度）
https://leetcode-cn.com/problems/longest-common-prefix/
class Solution {
public String longestCommonPrefix(String[] strs) {
if(strs.length == 0)
return "";
String ans = strs[0];
for(int i =1;i时间击败100%的算法
class Solution {
public String longestCommonPrefix(String[] strs) {
if (strs.length == 0){
return "";
}
String target = strs[0];
for (int i = 1; i < strs.length; i++) {
while (!strs[i].startsWith(target)){
target = target.substring(0,target.length()-1);
if (target.length()==0){
return "";
}
if (strs[i].startsWith(target)){
break;
}
}
}
return target;
}
}字符串中的单词数（simple难度）
https://leetcode-cn.com/problems/number-of-segments-in-a-string/
<方法一>：单指针
class Solution {
public int countSegments(String s) {
int segmentCount = 0;
for (int i = 0; i < s.length(); i++) {
if ((i == 0 || s.charAt(i-1) == ' ') && s.charAt(i) != ' ') {
segmentCount++;
}
}
return segmentCount;
}
}
作者：LeetCode
链接：https://leetcode-cn.com/problems/number-of-segments-in-a-string/solution/zi-fu-chuan-zhong-de-dan-ci-shu-by-leetcode/
来源：力扣（LeetCode）<方法二>：双指针
class Solution {
public int countSegments(String s) {
//去除掉前后空格，力扣有测试用例：" "
s = s.trim();
//长度为0或者1直接返回
if(s.length() == 0 ||s.length() == 1){
return s.length();
}
int res = 0;
int left = 0;
int right = s.length() - 1;
while (left < right) {
if (s.charAt(left) ==' ' && s.charAt(left - 1) !=' ') {
res++;
}
left++;
if (s.charAt(right) ==' ' && s.charAt(right + 1) !=' ') {
res++;
}
right--;
}
//此时left == right;如果left和right正在指向同一个字符
//例如："abc abc abc"，此时left和right指向第二个abc的b
//此时指向的字符不为空，因为第二个abc没有单独被left或right指针遍历完
//所以res没有记录第二个abc，所以此时res要加1
if (s.charAt(left) != ' ') {
res++;
//例如"abc abcd"
//此时left指向空字符，right指向a,
//但此时left刚刚遍历完的单词没有经过判断
//res没有记录，此时要判断一下
}else if(s.charAt(left) == ' ' && s.charAt(left - 1) != ' '){
res++;
//同上，例如"abcd abc"
}else if(s.charAt(right) == ' ' && s.charAt(right + 1) != ' ') {
res++;
}
return res;
}
}
作者：HIT_whc
链接：https://leetcode-cn.com/problems/number-of-segments-in-a-string/solution/javashuang-zhi-zhen-shi-jian-ji-bai-100-fzdof/
来源：力扣（LeetCode）最后一个单词的长度（simple难度）
https://leetcode-cn.com/problems/length-of-last-word/
本题方法和代码来源：
作者：guanpengchn
链接：https://leetcode-cn.com/problems/length-of-last-word/solution/hua-jie-suan-fa-58-zui-hou-yi-ge-dan-ci-de-chang-d/
来源：力扣（LeetCode）
class Solution {
public int lengthOfLastWord(String s) {
int end = s.length() - 1;
while(end >= 0 && s.charAt(end) == ' ') end--;
if(end < 0) return 0;
int start = end;
while(start >= 0 && s.charAt(start) != ' ') start--;
return end - start;
}
}
作者：guanpengchn
链接：https://leetcode-cn.com/problems/length-of-last-word/solution/hua-jie-suan-fa-58-zui-hou-yi-ge-dan-ci-de-chang-d/
来源：力扣（LeetCode）字符串中的第一个唯一的字符（simple难度）
https://leetcode-cn.com/problems/first-unique-character-in-a-string/
本题方法和代码来源：
作者：sdwwld
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/javaduo-chong-fang-shi-jie-jue-by-sdwwld-iojm/
来源：力扣（LeetCode）
public int firstUniqChar(String s) {
int count[] = new int[26];
char[] chars = s.toCharArray();
//先统计每个字符出现的次数
for (int i = 0; i < s.length(); i++)
count[chars[i] - 'a']++;
//然后在遍历字符串s中的字符，如果出现次数是1就直接返回
for (int i = 0; i < s.length(); i++)
if (count[chars[i] - 'a'] == 1)
return i;
return -1;
}
作者：sdwwld
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/javaduo-chong-fang-shi-jie-jue-by-sdwwld-iojm/
来源：力扣（LeetCode）
public int firstUniqChar(String s) {
Map map = new HashMap();
char[] chars = s.toCharArray();
//先统计每个字符的数量
for (char ch : chars) {
map.put(ch, map.getOrDefault(ch, 0) + 1);
}
//然后在遍历字符串s中的字符，如果出现次数是1就直接返回
for (int i = 0; i < s.length(); i++) {
if (map.get(chars[i]) == 1) {
return i;
}
}
return -1;
}
作者：sdwwld
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/javaduo-chong-fang-shi-jie-jue-by-sdwwld-iojm/
来源：力扣（LeetCode） 
public int firstUniqChar(String s) {
for (int i = 0; i < s.length(); i++)
if (s.indexOf(s.charAt(i)) == s.lastIndexOf(s.charAt(i)))
return i;
return -1;
}
作者：sdwwld
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/javaduo-chong-fang-shi-jie-jue-by-sdwwld-iojm/
来源：力扣（LeetCode）找不同（simple难度）
https://leetcode-cn.com/problems/find-the-difference/
本题方法和代码来源：
作者：sweetiee
链接：https://leetcode-cn.com/problems/find-the-difference/solution/yi-ju-hua-zhao-bu-tong-reduce-gao-qi-lai-eqok/
来源：力扣（LeetCode）
class Solution {
public char findTheDifference(String s, String t) {
char res = 0;
for (char c: s.toCharArray()) {
res ^= c;
}
for (char c: t.toCharArray()) {
res ^= c;
}
return res;
}
}
作者：sweetiee
链接：https://leetcode-cn.com/problems/find-the-difference/solution/yi-ju-hua-zhao-bu-tong-reduce-gao-qi-lai-eqok/
来源：力扣（LeetCode）java炫技：
class Solution {
public char findTheDifference(String s, String t) {
return (char)(s + t).chars().reduce(0, (a, b) -> a ^ b);
}
}
作者：sweetiee
链接：https://leetcode-cn.com/problems/find-the-difference/solution/yi-ju-hua-zhao-bu-tong-reduce-gao-qi-lai-eqok/
来源：力扣（LeetCode）赎金信（simple难度）
https://leetcode-cn.com/problems/ransom-note/
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
if (ransomNote.length() == 0) {
return true;
}
if (magazine.length() == 0) {
return false;
}
int[] count = new int[26];
char[] magazineArray = magazine.toCharArray();
//统计magazine中各个字符出现的次数并放在数组中
for (int i = 0; i < magazineArray.length; i++) {
count[magazineArray[i] - 'a']++;
}
char[] ransomNoteArray = ransomNote.toCharArray();
//判断magazine中各个字符的个数能否满足ransomNote的需要
for (int i = 0; i < ransomNoteArray.length; i++) {
//如果当前ransomNote需要的字符已经被用完，则返回false
if (count[ransomNoteArray[i] - 'a'] == 0) {
return false;
} else {
count[ransomNoteArray[i] - 'a']--;
}
}
return true;
}
}
作者：HIT_whc
链接：https://leetcode-cn.com/problems/ransom-note/solution/javashu-zu-jie-fa-by-hit_whc-x3id/
来源：力扣（LeetCode）有效的字母异位词（simple难度）
https://leetcode-cn.com/problems/valid-anagram/
本题方法和代码来源：
作者：guanpengchn
链接：https://leetcode-cn.com/problems/valid-anagram/solution/hua-jie-suan-fa-242-you-xiao-de-zi-mu-yi-wei-ci-by/
来源：力扣（LeetCode）
class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length())
return false;
int[] alpha = new int[26];
for(int i = 0; i< s.length(); i++) {
alpha[s.charAt(i) - 'a'] ++;
alpha[t.charAt(i) - 'a'] --;
}
for(int i=0;i<26;i++)
if(alpha[i] != 0)
return false;
return true;
}
}
作者：guanpengchn
链接：https://leetcode-cn.com/problems/valid-anagram/solution/hua-jie-suan-fa-242-you-xiao-de-zi-mu-yi-wei-ci-by/
来源：力扣（LeetCode）
class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) {
return false;
}
Map table = new HashMap();
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
table.put(ch, table.getOrDefault(ch, 0) + 1);
}
for (int i = 0; i < t.length(); i++) {
char ch = t.charAt(i);
table.put(ch, table.getOrDefault(ch, 0) - 1);
if (table.get(ch) < 0) {
return false;
}
}
return true;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/valid-anagram/solution/you-xiao-de-zi-mu-yi-wei-ci-by-leetcode-solution/
来源：力扣（LeetCode） 
排序 ：（最快的方法）
class Solution {
public boolean isAnagram(String s, String t) {
if(s.length()!=t.length())
return false;
char [] sArray=s.toCharArray();
char [] tArray=t.toCharArray();
Arrays.sort(sArray);
Arrays.sort(tArray);
String ss=String.valueOf(sArray);
String tt=String.valueOf(tArray);
return ss.equals(tt);
}
}从英文中重建数字（medium难度）
https://leetcode-cn.com/problems/reconstruct-original-digits-from-english/
本题方法和代码来源：
作者：romantic-tesla
链接：https://leetcode-cn.com/problems/reconstruct-original-digits-from-english/solution/zhao-dao-shu-zi-dui-ying-de-ying-wen-zi-kdxkz/
来源：力扣（LeetCode）
class Solution {
public String originalDigits(String s) {
if (s == null || s.length() == 0) {
return s;
}
int[] num = new int[10];
int len = s.length();
int[] alpha = new int[26];
for (int i = 0; i < len; i++) {
alpha[s.charAt(i) - 'a']++;
}
num[0] = alpha['z' - 'a'];
num[2] = alpha['w' - 'a'];
num[4] = alpha['u' - 'a'];
num[6] = alpha['x' - 'a'];
num[8] = alpha['g' - 'a'];
num[5] = alpha['f' - 'a'] - num[4];
num[3] = alpha['h' - 'a'] - num[8];
num[7] = alpha['s' - 'a'] - num[6];
num[9] = alpha['i' - 'a'] - num[5] - num[6] - num[8];
num[1] = alpha['n' - 'a'] - num[7] - 2 * num[9];
StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < 10; i++) {
int count = num[i];
for (int j = 0; j < count; j++) {
stringBuilder.append((char) ('0' + i));
}
}
return stringBuilder.toString();
}
}
作者：romantic-tesla
链接：https://leetcode-cn.com/problems/reconstruct-original-digits-from-english/solution/zhao-dao-shu-zi-dui-ying-de-ying-wen-zi-kdxkz/
来源：力扣（LeetCode）猜数字游戏（medium难度）
https://leetcode-cn.com/problems/bulls-and-cows/
本题方法和代码来源：
作者：zyxwmj
链接：https://leetcode-cn.com/problems/bulls-and-cows/solution/yi-ci-bian-li-1-ms-10000-by-zyxwmj-d5sa/
来源：力扣（LeetCode）
class Solution {
public String getHint(String secret, String guess) {
// secretArray 和 guessArray 分别记录 两个字符串中非公牛的各个数字的数量
int[] secretArray = new int[10];
int[] guessArray = new int[10];
// 公牛
int A = 0;
for (int i = 0; i < secret.length(); i++) {
// 如果同位的数字相等则，公牛++
if (secret.charAt(i) == guess.charAt(i)) {
A++;
} else {
secretArray[secret.charAt(i) - '0']++;
guessArray[guess.charAt(i) - '0']++;
}
}
// 奶牛
int B = 0;
for (int i = 0; i < 10; i++) {
// 不同位上的相同数字的数量
B += Math.min(secretArray[i], guessArray[i]);
}
StringBuilder stringBuilder = new StringBuilder();
return stringBuilder.append(A).append('A').append(B).append('B').toString();
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/bulls-and-cows/solution/yi-ci-bian-li-1-ms-10000-by-zyxwmj-d5sa/
来源：力扣（LeetCode）
class Solution {
public String getHint(String secret, String guess) {
int[] array = new int[10];
int A = 0, B = 0;
for (int i = 0; i < secret.length(); i++) {
if (secret.charAt(i) == guess.charAt(i)) {
A++;
} else {
// 判断 guess 在 i 之前是否该数字
if (array[secret.charAt(i) - '0']++ < 0) {
B++;
}
// 判断 secret 在 i 之前是否该数字
if (array[guess.charAt(i) - '0']-- > 0) {
B++;
}
}
}
StringBuilder stringBuilder = new StringBuilder();
return stringBuilder.append(A).append('A').append(B).append('B').toString();
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/bulls-and-cows/solution/yi-ci-bian-li-1-ms-10000-by-zyxwmj-d5sa/
来源：力扣（LeetCode）Fizz Buzz（simple难度）
https://leetcode-cn.com/problems/fizz-buzz/
class Solution {
public List fizzBuzz(int n) {
// ans list
List ans = new ArrayList();
for (int num = 1; num <= n; num++) {
boolean divisibleBy3 = (num % 3 == 0);
boolean divisibleBy5 = (num % 5 == 0);
if (divisibleBy3 && divisibleBy5) {
// Divides by both 3 and 5, add FizzBuzz
ans.add("FizzBuzz");
} else if (divisibleBy3) {
// Divides by 3, add Fizz
ans.add("Fizz");
} else if (divisibleBy5) {
// Divides by 5, add Buzz
ans.add("Buzz");
} else {
// Not divisible by 3 or 5, add the number
ans.add(Integer.toString(num));
}
}
return ans;
}
}
作者：LeetCode
链接：https://leetcode-cn.com/problems/fizz-buzz/solution/fizz-buzz-by-leetcode/
来源：力扣（LeetCode） 
class Solution {
public List fizzBuzz(int n) {
// ans list
List ans = new ArrayList();
for (int num = 1; num <= n; num++) {
boolean divisibleBy3 = (num % 3 == 0);
boolean divisibleBy5 = (num % 5 == 0);
String numAnsStr = "";
if (divisibleBy3) {
// Divides by 3, add Fizz
numAnsStr += "Fizz";
}
if (divisibleBy5) {
// Divides by 5, add Buzz
numAnsStr += "Buzz";
}
if (numAnsStr.equals("")) {
// Not divisible by 3 or 5, add the number
numAnsStr += Integer.toString(num);
}
// Append the current answer str to the ans list
ans.add(numAnsStr);
}
return ans;
}
}
作者：LeetCode
链接：https://leetcode-cn.com/problems/fizz-buzz/solution/fizz-buzz-by-leetcode/
来源：力扣（LeetCode） 
class Solution {
public List fizzBuzz(int n) {
// ans list
List ans = new ArrayList();
// Hash map to store all fizzbuzz mappings.
HashMap fizzBizzDict =
new HashMap() {
{
put(3, "Fizz");
put(5, "Buzz");
}
};
for (int num = 1; num <= n; num++) {
String numAnsStr = "";
for (Integer key : fizzBizzDict.keySet()) {
// If the num is divisible by key,
// then add the corresponding string mapping to current numAnsStr
if (num % key == 0) {
numAnsStr += fizzBizzDict.get(key);
}
}
if (numAnsStr.equals("")) {
// Not divisible by 3 or 5, add the number
numAnsStr += Integer.toString(num);
}
// Append the current answer str to the ans list
ans.add(numAnsStr);
}
return ans;
}
}
作者：LeetCode
链接：https://leetcode-cn.com/problems/fizz-buzz/solution/fizz-buzz-by-leetcode/
来源：力扣（LeetCode） 
相对名次（simple难度）
https://leetcode-cn.com/problems/relative-ranks/
本题方法和代码来源：
作者：zyxwmj
链接：https://leetcode-cn.com/problems/relative-ranks/solution/san-chong-jie-fa-ji-shu-pai-xu-2-ms10000-ml6v/
来源：力扣（LeetCode）
class Solution {
public String[] findRelativeRanks(int[] nums) {
int n = nums.length;
int[] array = new int[n];
// 拷贝数组
System.arraycopy(nums, 0, array, 0, n);
// 对数组进行排序
Arrays.sort(array);
String[] result = new String[n];
for (int i = 0; i < n; i++) {
// 查找当前成绩排第几名
int index = n - Arrays.binarySearch(array, nums[i]);
switch (index) {
case 1:
result[i] = "Gold Medal";
break;
case 2:
result[i] = "Silver Medal";
break;
case 3:
result[i] = "Bronze Medal";
break;
default:
result[i] = String.valueOf(index);
}
}
return result;
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/relative-ranks/solution/san-chong-jie-fa-ji-shu-pai-xu-2-ms10000-ml6v/
来源：力扣（LeetCode）
class Solution {
public String[] findRelativeRanks(int[] nums) {
int n = nums.length;
String[] result = new String[n];
int max = 0;
// 找出找出最高的成绩
for (int num : nums) {
if (max < num) {
max = num;
}
}
// 下标为成绩，值为成绩在 nums 数组的下标
int[] array = new int[max + 1];
for (int i = 0; i < n; i++) {
array[nums[i]] = i + 1;
}
// 记录当前成绩的排名
int count = 1;
for (int i = array.length - 1; i >= 0; i--) {
if (array[i] != 0) {
// 根据排名进行赋值
switch (count) {
case 1:
result[array[i] - 1] = "Gold Medal";
break;
case 2:
result[array[i] - 1] = "Silver Medal";
break;
case 3:
result[array[i] - 1] = "Bronze Medal";
break;
default:
result[array[i] - 1] = String.valueOf(count);
}
count++;
}
}
return result;
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/relative-ranks/solution/san-chong-jie-fa-ji-shu-pai-xu-2-ms10000-ml6v/
来源：力扣（LeetCode）外观数列（simple难度）
https://leetcode-cn.com/problems/count-and-say/
本题方法和代码来源：
作者：zyxwmj
链接：https://leetcode-cn.com/problems/count-and-say/solution/xun-huan-he-di-gui-liang-chong-jie-fa-di-oof8/
来源：力扣（LeetCode）
class Solution {
public String countAndSay(int n) {
// 递归终止条件
if (n == 1) {
return "1";
}
// 获取到上一层的字符串
String s = countAndSay(n - 1);
StringBuilder result = new StringBuilder();
// 记录每个数字的开始索引
int start = 0;
for (int i = 1; i < s.length(); i++) {
// 当数字发生改变时执行
if (s.charAt(i) != s.charAt(start)) {
result.append(i - start).append(s.charAt(start));
start = i;
}
}
// 字符串最后一个数字
result.append(s.length() - start).append(s.charAt(start));
return result.toString();
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/count-and-say/solution/xun-huan-he-di-gui-liang-chong-jie-fa-di-oof8/
来源：力扣（LeetCode）压缩字符串（medium难度）
https://leetcode-cn.com/problems/string-compression/
class Solution {
public int compress(char[] chars) {
int anchor = 0, write = 0;
for (int read = 0; read < chars.length; read++) {
if (read + 1 == chars.length || chars[read + 1] != chars[read]) {
chars[write++] = chars[anchor];
if (read > anchor) {
for (char c: ("" + (read - anchor + 1)).toCharArray()) {
chars[write++] = c;
}
}
anchor = read + 1;
}
}
return write;
}
}
作者：LeetCode
链接：https://leetcode-cn.com/problems/string-compression/solution/ya-suo-zi-fu-chuan-by-leetcode/
来源：力扣（LeetCode）判断子序列（simple难度）
https://leetcode-cn.com/problems/is-subsequence/
class Solution {
public boolean isSubsequence(String s, String t) {
int n = s.length(), m = t.length();
int i = 0, j = 0;
while (i < n && j < m) {
if (s.charAt(i) == t.charAt(j)) {
i++;
}
j++;
}
return i == n;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/is-subsequence/solution/pan-duan-zi-xu-lie-by-leetcode-solution/
来源：力扣（LeetCode）
class Solution {
public boolean isSubsequence(String s, String t) {
int n = s.length(), m = t.length();
int[][] f = new int[m + 1][26];
for (int i = 0; i < 26; i++) {
f[m][i] = m;
}
for (int i = m - 1; i >= 0; i--) {
for (int j = 0; j < 26; j++) {
if (t.charAt(i) == j + 'a')
f[i][j] = i;
else
f[i][j] = f[i + 1][j];
}
}
int add = 0;
for (int i = 0; i < n; i++) {
if (f[add][s.charAt(i) - 'a'] == m) {
return false;
}
add = f[add][s.charAt(i) - 'a'] + 1;
}
return true;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/is-subsequence/solution/pan-duan-zi-xu-lie-by-leetcode-solution/
来源：力扣（LeetCode）
通过删除字母匹配到字典里最长的单词（medium难度）
https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/
本题方法和代码来源：
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode）
class Solution {
public boolean isSubsequence(String t, String s) {
int indext = 0, indexs = 0;
while (indext < t.length() && indexs < s.length()) {
if (t.charAt(indext) == s.charAt(indexs)) {
indext++;
}
indexs++;
}
return indext == t.length();
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode）
class Solution {
public boolean isSubsequence(String t, String s) {
int index = -1;
for (int i = 0; i < t.length(); i++) {
index = s.indexOf(t.charAt(i), index + 1);
if (index == -1) {
return false;
}
}
return true;
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode）
class Solution {
public boolean isSubsequence(String t, String s) {
int n = s.length();
int[][] dp = new int[n + 1][26];
// 初始化结尾
for (int i = 0; i < 26; i++) {
dp[n][i] = n;
}
// 初始化 dp 数组
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j < 26; j++) {
if (s.charAt(i) - 'a' == j) {
dp[i][j] = i;
} else {
// 不相等，则继承上一位的值
dp[i][j] = dp[i + 1][j];
}
}
}
int index = 0;
for (char c : t.toCharArray()) {
// 如果下一个字符在结尾，则表示 s 中没有该字符
if (dp[index][c - 'a'] == n) {
return false;
}
// 移动到下一个字符的后面
index = dp[index][c - 'a'] + 1;
}
return true;
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode）
class Solution {
public String findLongestWord(String s, List d) {
String result = "";
for (String t : d) {
if (isSubsequence(t, s)) {
// 获取长度最长且字典顺序最小的字符串
if (result.length() < t.length() || (result.length() == t.length() && result.compareTo(t) > 0)) {
result = t;
}
}
}
return result;
}
// 判断 t 是否为 s 的子序列
public boolean isSubsequence(String t, String s) {
int indext = 0, indexs = 0;
while (indext < t.length() && indexs < s.length()) {
if (t.charAt(indext) == s.charAt(indexs)) {
indext++;
}
indexs++;
}
return indext == t.length();
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode） 
class Solution {
public String findLongestWord(String s, List d) {
String result = "";
for (String t : d) {
if (isSubsequence(t, s)) {
// 获取长度最长且字典顺序最小的字符串
if (result.length() < t.length() || (result.length() == t.length() && result.compareTo(t) > 0)) {
result = t;
}
}
}
return result;
}
// 判断 t 是否为 s 的子序列
public boolean isSubsequence(String t, String s) {
int index = -1;
for (int i = 0; i < t.length(); i++) {
index = s.indexOf(t.charAt(i), index + 1);
if (index == -1) {
return false;
}
}
return true;
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode 
class Solution {
public String findLongestWord(String s, List d) {
int n = s.length();
int[][] dp = new int[n + 1][26];
// 初始化结尾
for (int i = 0; i < 26; i++) {
dp[n][i] = n;
}
// 初始化 dp 数组
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j < 26; j++) {
if (s.charAt(i) - 'a' == j) {
dp[i][j] = i;
} else {
dp[i][j] = dp[i + 1][j];
}
}
}
String result = "";
for (String t : d) {
// count 记录相等字符的个数
int count = 0, index = 0;
for (char c : t.toCharArray()) {
// 如果下一个字符的下标为 n，则表示该字符不存在
if (dp[index][c - 'a'] == n) {
break;
}
count++;
// 移动到下一个字符的后面
index = dp[index][c - 'a'] + 1;
}
// 判断是否到了最后一个字符
if (count == t.length()) {
// 长度最长且字典顺序最小的字符串
if (t.length() > result.length() || (t.length() == result.length() && result.compareTo(t) > 0)) {
result = t;
}
}
}
return result;
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/fei-jing-xin-si-xie-de-dong-tai-gui-hua-xu2o9/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。 加一（simple难度）
https://leetcode-cn.com/problems/plus-one/
本题方法和代码来源：
作者：yhhzw
链接：https://leetcode-cn.com/problems/plus-one/solution/java-shu-xue-jie-ti-by-yhhzw/
来源：力扣（LeetCode）
class Solution {
public int[] plusOne(int[] digits) {
for (int i = digits.length - 1; i >= 0; i--) {
digits[i]++;
digits[i] = digits[i] % 10;
if (digits[i] != 0) return digits;
}
digits = new int[digits.length + 1];
digits[0] = 1;
return digits;
}
}
作者：yhhzw
链接：https://leetcode-cn.com/problems/plus-one/solution/java-shu-xue-jie-ti-by-yhhzw/
来源：力扣（LeetCode）二进制求和（simple难度）
https://leetcode-cn.com/problems/add-binary/
本题方法和代码来源：
作者：guanpengchn
链接：https://leetcode-cn.com/problems/add-binary/solution/hua-jie-suan-fa-67-er-jin-zhi-qiu-he-by-guanpengch/
来源：力扣（LeetCode）
class Solution {
public String addBinary(String a, String b) {
StringBuilder ans = new StringBuilder();
int ca = 0;
for(int i = a.length() - 1, j = b.length() - 1;i >= 0 || j >= 0; i--, j--) {
int sum = ca;
sum += i >= 0 ? a.charAt(i) - '0' : 0;
sum += j >= 0 ? b.charAt(j) - '0' : 0;
ans.append(sum % 2);
ca = sum / 2;
}
ans.append(ca == 1 ? ca : "");
return ans.reverse().toString();
}
}
作者：guanpengchn
链接：https://leetcode-cn.com/problems/add-binary/solution/hua-jie-suan-fa-67-er-jin-zhi-qiu-he-by-guanpengch/
来源：力扣（LeetCode）
字符串相加（simple难度）
https://leetcode-cn.com/problems/add-strings/
class Solution {
public String addStrings(String num1, String num2) {
StringBuilder res = new StringBuilder();
int ca = 0;
for (int i = num1.length()-1, j = num2.length()-1; i >= 0 || j >= 0; i--, j--) {
int sum = ca;
sum += i >= 0 ? num1.charAt(i) - '0' : 0;
sum += j >= 0 ? num2.charAt(j) - '0' : 0;
res.append(sum % 10);
ca = sum / 10;
}
res.append(ca == 1 ? ca : "");
return res.reverse().toString();
}
}密钥格式化（simple难度）
https://leetcode-cn.com/problems/license-key-formatting/
本题方法和代码来源：
作者：zyxwmj
链接：https://leetcode-cn.com/problems/license-key-formatting/solution/liang-chong-jie-fa-shi-yong-char-4-ms990-xrpo/
来源：力扣（LeetCode）
class Solution {
public String licenseKeyFormatting(String S, int K) {
StringBuilder s = new StringBuilder();
// 统计已打印字符个数
int count = 0;
// 倒序遍历字符串
for (int i = S.length() - 1; i >= 0; i--) {
if (S.charAt(i) != '-') {
// 计算什么时候打印分隔符
if (count != 0 && count % K == 0) {
s.append('-');
}
// 转为大写字母添加
s.append(Character.toUpperCase(S.charAt(i)));
count++;
}
}
// 反转字符串
return s.reverse().toString();
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/license-key-formatting/solution/liang-chong-jie-fa-shi-yong-char-4-ms990-xrpo/
来源：力扣（LeetCode）
class Solution {
public String licenseKeyFormatting(String S, int K) {
char[] array = S.toCharArray();
// 统计字符的个数
int count = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] != '-') {
// 将小写字符转为大写字符
if (array[i] > 'Z') {
array[i] -= 32;
}
count++;
}
}
// 没有字符返回空串
if (count == 0) {
return "";
}
// 计算分隔符的个数
int separator = count / K;
// 如果正好整除，分隔符个数 - 1
separator = count % K == 0 ? separator - 1 : separator;
// 存储结果的数组
char[] result = new char[count + separator];
// 指向结果数组的下标，倒序赋值
int index = result.length - 1;
// 统计已打印字符个数
int letter = 0;
for (int i = S.length() - 1; i >= 0; i--) {
if (array[i] != '-') {
// 计算什么时候打印分隔符
if (letter != 0 && letter % K == 0) {
result[index--] = '-';
}
result[index--] = array[i];
letter++;
}
}
return new String(result);
}
}
作者：zyxwmj
链接：https://leetcode-cn.com/problems/license-key-formatting/solution/liang-chong-jie-fa-shi-yong-char-4-ms990-xrpo/
来源：力扣（LeetCode）Z字形变换（medium难度）
https://leetcode-cn.com/problems/zigzag-conversion/
本题方法和代码来源：
作者：jyd
链接：https://leetcode-cn.com/problems/zigzag-conversion/solution/zzi-xing-bian-huan-by-jyd/
来源：力扣（LeetCode）
class Solution {
public String convert(String s, int numRows) {
if(numRows < 2) return s;
List rows = new ArrayList();
for(int i = 0; i < numRows; i++) rows.add(new StringBuilder());
int i = 0, flag = -1;
for(char c : s.toCharArray()) {
rows.get(i).append(c);
if(i == 0 || i == numRows -1) flag = - flag;
i += flag;
}
StringBuilder res = new StringBuilder();
for(StringBuilder row : rows) res.append(row);
return res.toString();
}
}
作者：jyd
链接：https://leetcode-cn.com/problems/zigzag-conversion/solution/zzi-xing-bian-huan-by-jyd/
来源：力扣（LeetCode） 实现strStr()（simple难度）
https://leetcode-cn.com/problems/implement-strstr/
