leetcode 系列
6. ZigZag Conversion,将字符串按Z字形排列并横向输出
public class Solution {public String convert ( String s , int nRows ) {int len = s . length ();if ( len == 0 || nRows <= 1 ) return s ;String [] ans = new String [ nRows ];Arrays . fill ( ans , "" );int row = 0 , delta = 1 ;for ( int i = 0 ; i < len ; i ++) {ans [ row ] += s . charAt ( i );row += delta ;if ( row >= nRows ) {row = nRows - 2 ;delta = - 1 ;}if ( row < 0 ) {row = 1 ;delta = 1 ;}}String ret = "" ;for ( int i = 0 ; i < nRows ; i ++) {ret += ans [ i ];}return ret ;}}
7 inverse integer
这道题是反转输出整数，需要考虑溢出的问题
首先我们要清楚int的范围int -2147483648～2147483647
如果反转后超出这个范围 我们就要返回-1
public class Solution {
public static int reverse(int x)
{
char[] numArray = null;
String numStr="";
if (x == 0) {
return 0;
} else if (x > 0) {
numArray = (x + "").toCharArray();
} else {
numStr+="-";
numArray = (x + "").substring(1).toCharArray();
}
for(int i =numArray.length-1;i>=0;i--){
numStr+=numArray[i];
}
long res=Long.parseLong(numStr);
if(res<-2147483648||res>2147483647)
return 0;
else
{
return (int)res;
}
}
}
10 10. Regular Expression Matching
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true
<span style="font-size:14px;">public class Solution
{
public boolean isMatch(String s, String p)
{
if(p.length() == 0)
return s.length() == 0;
//p's length 1 is special case
if(p.length() == 1 || p.charAt(1) != '*')
{
if(s.length() < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0)))
return false;
return isMatch(s.substring(1), p.substring(1));
}
else
{
int len = s.length();
int i = -1;
while(i<len && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){
if(isMatch(s.substring(i+1), p.substring(2)))
return true;
i++;
}
return false;
}
}
}</span>
27. Remove Element
<span style="font-size:14px;">public class Solution {
public int removeElement(int[] nums, int val)
{ int i=0;
int j=0;
while(j < nums.length){
if(nums[j] != val){
nums[i] = nums[j];
i++;
}
j++;
}
return i;
}
}</span>
20. Valid Parentheses
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
package leetcode;
import java.util.Stack;
public class Le20
{
public static boolean isValid(String s)
{
char[] ca=s.toCharArray();
if(s.length()==0)
return true;
if(s.length()==1)
{
if(ca[0]!='('&&ca[0]!=')'&&ca[0]!='['&&ca[0]!=']'&&ca[0]!='['&&ca[0]!='{'&&ca[0]!='}')
return true;
else
return false;
}
Stack stack=new Stack();
for(int i=0;i<ca.length;i++)
{
if(ca[i]=='('||ca[i]==')'||ca[i]=='['||ca[i]==']'||ca[i]=='['||ca[i]=='{'||ca[i]=='}')
{
if(stack.isEmpty())
{
stack.push(ca[i]);
}
else
{
char tmp=(char)stack.peek();
if((tmp=='('&&ca[i]==')')||(tmp=='['&&ca[i]==']')||(tmp=='{'&&ca[i]=='}'))
{
stack.pop();
}
else
{
stack.push(ca[i]);
}
}
}
}
if(stack.size()==0)
{
return true;
}
else
{
return false;
}
}
public static void main(String[] args)
{
boolean flag=isValid("a()b");
System.out.println(flag);
}
}
127. Word Ladder
Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord to endWord, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the word list
For example,
Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Note:
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
public int ladderLength(String beginWord, String endWord, Set<String> wordList)
{
wordList.add(endWord);
if(wordList.contains("beginWord"))
{
wordList.remove("beginWord");
}
Queue<String> que = new LinkedList<String>();
ArrayList<String> remove=new ArrayList<String>();
HashMap<String,Integer> hashmap=new HashMap<String,Integer>();
HashMap<String,ArrayList<String>> result=new HashMap<String,ArrayList<String>>();
hashmap.put(beginWord, 1);
que.add(beginWord);
int floor=1;
while(!que.isEmpty())
{
String tmp=que.poll();
result.put(tmp, new ArrayList<String>());
int cur=hashmap.get(tmp);
System.out.println("第几层？"+cur+"正在处理的string为："+tmp);
if(tmp.equals(endWord))
{
System.out.println("这就是答案"+tmp+" "+cur);
return cur;
}
if(cur!=floor)
{
System.out.println("这是新的一层");
for(int i=0;i<remove.size();i++)
{
wordList.remove(remove.get(i));
System.out.println("移除set中上一层的元素:"+remove.get(i));
}
floor=cur;
System.out.println("更新现在正在访问的层数floor："+floor);
remove.clear();
System.out.println("清除所有remove内容");
}
for(String s:wordList)
{
if(near(tmp,s))
{
System.out.println(tmp+"可以经过一步变为："+s);
ArrayList<String> list=new ArrayList<String>();
System.out.println("新建链表 为期赋值 然后加入到 上一个结点的list中");
result.get(tmp).add(s);
que.add(s);
System.out.println("队列中加入新的结点"+s);
remove.add(s);
System.out.println("需要移除的string："+s);
hashmap.put(s, hashmap.get(tmp)+1);
System.out.println("为此结点的深度加一 "+s+" "+hashmap.get(tmp));
}
}
}
return 0;
}
public static boolean near(String s1,String s2)
{
if(s1.length()!=s2.length())
{
return false;
}
int len1=s1.length();
int len2=s2.length();
int count=0;
for(int i=0,j=0;i<s1.length();i++,j++)
{
if(s1.charAt(i)!=s2.charAt(j))
{
count++;
}
}
if(count==1)
{
return true;
}
return false;
}