131 Palindrome Partitioning [Leetcode]

题目内容：

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"]
]

题目分析：

使用回溯的方法做，也可以称为DFS的方法。先穷举所有回文分支下的每个分支所有的状况，再添加到结果中去，再回溯回上一个节点。

代码实现，时间8ms：

class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> result;
vector<string> temp;
backTracking(s, 0, result, temp);
return result;
}

void backTracking(string &s, int index, vector<vector<string>> &result, vector<string> &temp) {
if(index >= s.size()) {
result.push_back(temp);
return;
}

for(int i = index; i < s.size(); ++i) {
if(isPalindrome(index, i, s)) {
temp.push_back(s.substr(index, i-index+1));
backTracking(s, i+1, result, temp);
temp.pop_back();//backtrack
}
}
}

bool isPalindrome(int start, int end, string &s) {
while(start < end) {
if(s[start] != s[end])
return false;
++start;
--end;
}
return true;
}
};