LeetCode-301. Remove Invalid Parentheses

LeetCode-301. Remove Invalid Parentheses

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ​​(​​​ and ​​)​​.

Example 1:

Input: "()())()"Output: ["()()()", "(())()"]

Example 2:

Input: "(a)())()"Output: ["(a)()()", "(a())()"]

Example 3:

Input: ")("Output: [""]

题解：

使用dfs查找，先遍历字符串找出最少需要删除的左括号和右括号个数，然后dfs删除。

class Solution {public: bool isValid(string s) { int cnt = 0; for (int i = 0; i < s.length(); i++) { if (s[i] == '(') { cnt++; } else if (s[i] == ')') { cnt--; if (cnt < 0) { return false; } } } return cnt == 0; } void dfs(int k, string &s, unordered_set &res, string tmp, int left, int right, int n) { if (k == n) { if (left == 0 && right == 0 && isValid(tmp) == true) { res.insert(tmp); } } else { if (s[k] == '(') { if (left > 0) { dfs(k + 1, s, res, tmp, left - 1, right, n); } } else if (s[k] == ')') { if (right > 0) { dfs(k + 1, s, res, tmp, left, right - 1, n); } } dfs(k + 1, s, res, tmp + s[k], left, right, n); } } vector removeInvalidParentheses(string s) { int n = s.length(); if (n == 0) { return {""}; } int left = 0, right = 0; for (int i = 0; i < n; i++) { if (s[i] == '(') { left++; } if (s[i] == ')') { if (left > 0) { left--; } else { right++; } } } unordered_set res; dfs(0, s, res, "", left, right, n); return vector(res.begin(), res.end()); }};

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