【LeetCode】131. Palindrome Partitioning

131. Palindrome Partitioning

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• Difficulty: Medium
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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"]]

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分析

在每一步都可以判断中间结果是否为合法结果，用回溯法

一个长度为n的字符串，有n-1个地方可以砍断，每个地方可断可不断，因此时间复杂度为O(2^n)

class Solution {public:vector<vector<string>> partition(string s) {vector<vector<string>> result;vector<string> path; // ???? partition ????dfs(s, path, result, 0, 1);return result;}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;}dfs(s, path, result, prev, start + 1);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;}};