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：
本题考查回溯算法。从下标0开始遍历字符串，一旦在下标 i 找到回文子字符串，那么就把下标从 0 到 i 的子字符串加入temp中，继续从下标 i 接着往下找，一旦curIndex等于字符串长度，那么就把temp加入到result中。如果一直到最后都没找到回文子字符串，那么就进行剪枝。

class Solution {    public List<List<String>> partition(String s) {          List<List<String>> result = new ArrayList<List<String>>();          List<String> temp = new ArrayList<String>();          dfs(s, 0, temp, result);          return result;      }      private void dfs(String s, int curIndex, List<String> temp, List<List<String>> result) {          if (curIndex == s.length()) {              result.add(new ArrayList<String>(temp));              return;          }          for (int i = curIndex + 1; i <= s.length(); i++) {              String prefix = s.substring(curIndex, i);              if (!isPrlindrome(prefix))  //剪枝                  continue;              temp.add(prefix);              dfs(s, i, temp, result);              temp.remove(temp.size() - 1);          }      }      private boolean isPrlindrome(String s) {          int left = 0;          int right = s.length() - 1;          while (left < right) {              if (s.charAt(left) != s.charAt(right))                  return false;              left++;              right--;          }          return true;      }  }