LeetCode 131 Palindrome Partitioning

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

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [    ["aa","b"],    ["a","a","b"]  ]

思路：1.判断字符串的字串S.subString(i,j) [i<=j]是否为为回文子串,用boolean型的二维数组isPalindrome来存储该结果。在这个地方用了点小技巧，isPalindrome[i]j]会依赖于isPalindrome[i+1]j-1]  [i+2<=j].

          2.假设在求String s的所有回文子串，我们已经知道了s.substring(0,0),s.substring(0,1),s.substring(0,2),s.substring(0,3),....,s.substring(0,s.length()-2)的所有回文字串，那我们只需要遍历isPalindrome[i]js.lenth()-1]是不是true｛即从s.substring(i)是不是回文字符串｝若为true，我们取出s.substring(i-1)的所有回文字串，并在每一种可能性末尾添加s.substring(i)即可，代码如下

public class Solution {public boolean[][] isPalindrome(String s) {boolean[][] isPalindrome = new boolean[s.length()][s.length()];for (int i = 0; i < s.length(); i++)isPalindrome[i][i] = true;for (int i = 0; i < s.length() - 1; i++)isPalindrome[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));for (int length = 2; length < s.length(); length++) {for (int start = 0; start + length < s.length(); start++) {isPalindrome[start][start + length] = isPalindrome[start + 1][start+ length - 1]&& s.charAt(start) == s.charAt(start + length);}}return isPalindrome;}public List<List<String>> partition(String s) {boolean[][] isPalindrome= isPalindrome(s);HashMap<Integer,List<List<String>>> hm=new HashMap<Integer,List<List<String>>>();for(int i=0;i<s.length();i++){List<List<String>> ls=new ArrayList<List<String>>();if(isPalindrome[0][i]){ArrayList<String> temp=new ArrayList<String>();temp.add(s.substring(0, i+1));ls.add(temp);}for(int j=1;j<=i;j++){if(isPalindrome[j][i]){List<List<String>> l=hm.get(j-1);List<List<String>> al=new ArrayList<List<String>>();for(List<String> temp:l){ArrayList<String> clone=new ArrayList<String>(temp);clone.add(s.substring(j, i+1));al.add(clone);}ls.addAll(al);}}hm.put(i,ls);}return hm.get(s.length()-1);}}