131. Palindrome Partitioning 131.回文分区

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

Example:

Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]

1 图示：双重循环的逻辑

ajhdfhjfbhjrfbkef
  s i i+1 i+2...
    s i i+1 i+2...

2 递归的思想：start变成i+1, i相应往后。start的变化是在递归里完成的。

for(int i=start;i<s.length();i++){ 从变量循环
                if(isPal(s,start,i)){
                    list.add(s.substring(start,i+1)); 
                    dfs(s,i+1,list,res); start变成i+1, i相应往后。start的变化是在递归里完成的。
                    list.remove(list.size()-1);
                }
            }

3 不知道isPalindrome(String s, int low, int high)要这样写，更方便

4 length()要加括号

class Solution {
    public List<List<String>> partition(String s) {
        //cc
        List<String> temp = new ArrayList<>();
        List<List<String>> result = new ArrayList<>();
        
        if (s == null) return null;
        if (s.length() == 0) return result;
        
        dfs(s, 0, temp, result);
        
        return result;
    }
    
    public void dfs(String s, int start, List<String> temp, List<List<String>> result) {
        if (start == s.length()) {
            result.add(new ArrayList(temp));
        }else {
            for (int i = start; i < s.length(); i++) {
                if (isPalindrome(s, start, i)) {
                    temp.add(s.substring(start, i + 1));
                    dfs(s, i + 1, temp, result);
                    temp.remove(temp.size() - 1);
                }
            }
        }
    }
    
    //isPalindrome
    public boolean isPalindrome(String s, int start, int end) {
        while (start < end) {
            if (s.charAt(start++) != s.charAt(end--))
                return false;
        }
        return true;
    }
}