LeetCode | 344. Reverse String, 541. Reverse String II, 剑05. 替换空格,151. Reverse Words in a String,剑58
344. Reverse String
Link: https://leetcode.com/problems/reverse-string/
Description
Write a function that reverses a string. The input string is given as an array of characters s.
You must do this by modifying the input array in-place with O(1) extra memory.
Approach
Two pointers
- Initialize
startpoints to the first element ofs,endpoints to the last element ofs. - Swap
startandend, then both pointers move towards the middle at the same time. - Stop when
startandendmeet.
Solution
class Solution {
public void reverseString(char[] s) {
int left = 0;
int right = s.length - 1;
while (left < right) {
char temp = s[left];
s[left] = s[right];
s[right] = temp;
left++;
right--;
}
}
}
541. Reverse String II
Link: https://leetcode.com/problems/reverse-string-ii/
Description
Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.
If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.
Approach
Two pointers
- Traverse
s:- If there are more than
kcharacters left, reversekelements; else, reverse all of them - let
i = i + 2k, traverse every2kcharacters.
- If there are more than
Solution
class Solution {
public String reverseStr(String s, int k) {
char[] c = s.toCharArray();
for (int i = 0; i < c.length; i += 2 * k) {
if (i + k < c.length) {
swap(c, i, i + k - 1);
}
else {
swap(c, i, c.length - 1);
}
}
return new String(c);
}
private void swap(char[] c, int left, int right) {
while (left < right) {
char temp = c[left];
c[left] = c[right];
c[right] = temp;
left++;
right--;
}
}
}
class Solution {
public String reverseStr(String s, int k) {
StringBuffer result = new StringBuffer();
int start = 0;
while (start < s.length()) {
StringBuffer temp = new StringBuffer();
int firstK = start + k < s.length() ? start + k : s.length();
int secondK = start + 2 * k < s.length() ? start + 2 * k : s.length();
temp.append(s.substring(start, firstK));
result.append(temp.reverse());
if (firstK < secondK)
result.append(s.substring(firstK, secondK));
start += 2 * k;
}
return result.toString();
}
}
剑指 Offer 05. 替换空格
Description
请实现一个函数，把字符串 s 中的每个空格替换成"%20"。
Approach
Two pointers
- Expand the size of the array to the size after replacing each space with “%20”
- Initialize pointer
leftpoints to the end of the previous array,rightto the end of the expanded array. - Traverse from
left:- If
s[left]= ' ', replace it with “%20”; else lets[left] = s[right] left--, right--
- If
Solution
class Solution {
public String replaceSpace(String s) {
if (s == null || s.length() == 0)
return s;
StringBuilder sb = new StringBuilder();
for (char c: s.toCharArray()) {
if (c == ' ')
sb.append(" ");
}
int left = s.length() - 1;
s += sb.toString();
int right = s.length() - 1;
char[] ch = s.toCharArray();
while (left >= 0) {
if (ch[left] == ' ') {
ch[right--] = '0';
ch[right--] = '2';
ch[right] = '%';
}
else
ch[right] = ch[left];
right--;
left--;
}
return new String(ch);
}
}
151. Reverse Words in a String
Link: https://leetcode.com/problems/reverse-words-in-a-string/
Description
Given an input string s, reverse the order of the words.
A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.
Return a string of the words in reverse order concatenated by a single space.
Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.
Approach
Two pointers
- Remove redundant spaces:
- Remove the spaces at the start and spaces at the end.
- Remove the redundant spaces between the words:
- If the previous element of a space is a character, remove it
- Reverse the array
- Initialize two pointers
startandend - Swap start and end, then both pointers move towards the middle at the same time
- Stop when
startandendmeet.
- Initialize two pointers
- Reverse each word
- Initialize two pointers
startandend - Traverse the array:
- Let start points to the
startof a word, end points to theendof the word, which is the element before the space. - Reverse the word, then let
startandendpoints to the next word
- Let start points to the
- Initialize two pointers
Solution
class Solution {
public String reverseWords(String s) {
StringBuilder sb = removeSpace(s);
reverseString(sb, 0, sb.length() - 1);
return reverseEachWord(sb);
}
private StringBuilder removeSpace(String s) {
char[] ch = s.toCharArray();
int start = 0;
int end = s.length() - 1;
StringBuilder sb = new StringBuilder();
while (ch[start] == ' ')
start++;
while (ch[end] == ' ')
end--;
while (start <= end) {
char c = ch[start];
if (c != ' ' || sb.charAt(sb.length() - 1) != ' ')
sb.append(c);
start++;
}
return sb;
}
private void reverseString(StringBuilder sb, int start, int end) {
while (start < end) {
char temp = sb.charAt(start);
sb.setCharAt(start, sb.charAt(end));
sb.setCharAt(end, temp);
start++;
end--;
}
}
private String reverseEachWord(StringBuilder sb) {
int start = 0;
int end = 1;
while (start < sb.length()) {
while (end < sb.length() && sb.charAt(end) != ' ')
end++;
reverseString(sb, start, end - 1);
start = end + 1;
end = start + 1;
}
return sb.toString();
}
}
剑指Offer58-II.左旋转字符串
Link: https://leetcode.cn/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/
Description
字符串的左旋转操作是把字符串前面的若干个字符转移到字符串的尾部。请定义一个函数实现字符串左旋转操作的功能。比如，输入字符串"abcdefg"和数字2，该函数将返回左旋转两位得到的结果"cdefgab"。
Approach
- Reverse the first n character
- Reverse the remain characters
- Reverse the whole String
Solution
class Solution {
public String reverseLeftWords(String s, int n) {
StringBuilder sb= new StringBuilder(s);
reverseString(sb, 0, n - 1);
reverseString(sb, n, s.length() - 1);
return sb.reverse().toString();
}
private void reverseString(StringBuilder sb, int start, int end) {
while (start < end) {
char temp = sb.charAt(start);
sb.setCharAt(start, sb.charAt(end));
sb.setCharAt(end, temp);
start++;
end--;
}
}
}