【leetcode】回文串划分（DFS）

131. Palindrome Partitioning

leetcode题目

题目描述

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”]
]

题解

深度优先搜索，分两种情况，一种是不断开，一种是可断开时断开。

Solution1:

class Solution {public:    vector<vector<string>> partition(string s) {        vector<vector<string>> result;        vector<string> path;        dfs(s, path, result, 0, 1);        return result;    }    // prev 表示前一个隔板， start表示当前隔板    void dfs(string& s, vector<string>& path,             vector<vector<string>>& result, size_t prev, size_t start) {        if (start == s.size()) {            if (isPalindrome(s, prev, start-1)) {                path.push_back(s.substr(prev, start-prev));                result.push_back(path);                path.pop_back();                }            return;         }        // 不断开        dfs(s, path, result, prev, start+1);        // 断开        if (isPalindrome(s, prev, start-1)) {            path.push_back(s.substr(prev, start-prev));            dfs(s, path, result, start, start+1);            path.pop_back();                    }    }    bool isPalindrome(const string& s, int start, int end) {        while(start < end && s[start] == s[end]) {            ++start;            --end;        }        return start >= end;    }};