From 091918466d98a84bde5f7a7ffe4aea343bf77885 Mon Sep 17 00:00:00 2001 From: ShusenTang Date: Tue, 28 Apr 2020 23:17:52 +0800 Subject: [PATCH] add 301. Remove Invalid Parentheses :beer: --- README.md | 1 + solutions/301. Remove Invalid Parentheses.md | 131 +++++++++++++++++++ 2 files changed, 132 insertions(+) create mode 100644 solutions/301. Remove Invalid Parentheses.md diff --git a/README.md b/README.md index 893d196..935c495 100644 --- a/README.md +++ b/README.md @@ -221,6 +221,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。 | 295 |[Find Median from Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)|[C++](solutions/295.%20Find%20Median%20from%20Data%20Stream.md)|Hard| | | 297 |[Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)|[C++](solutions/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.md)|Hard| | | 300 |[Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/300.%20Longest%20Increasing%20Subsequence.md)|Medium| | +| 301 |[Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/)|[C++](solutions/301.%20Remove%20Invalid%20Parentheses.md)|Hard| | | 303 |[Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/303.%20Range%20Sum%20Query%20-%20Immutable.md)|Easy| | | 304 |[Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/304.%20Range%20Sum%20Query%202D%20-%20Immutable.md)|Medium| | | 306 |[Additive Number](https://leetcode.com/problems/additive-number/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/306.%20Additive%20Number.md)|Medium| | diff --git a/solutions/301. Remove Invalid Parentheses.md b/solutions/301. Remove Invalid Parentheses.md new file mode 100644 index 0000000..5ccd031 --- /dev/null +++ b/solutions/301. Remove Invalid Parentheses.md @@ -0,0 +1,131 @@ +# [301. Remove Invalid Parentheses](https://leetcode.com/problems/remove-invalid-parentheses/) + +# 思路 + +## 思路一、暴力BFS + +由于要求去掉最少的字符，所以考虑用BFS（不能DFS）：把给定字符串入队，然后取出检测其是否合法，若合法就将其存放到结果数组res；不合法的话，对其进行遍历，依次穷举去掉每个括号然后得到一个新串，如果这个字符串之前没有遇到过（所以需要用一个set来记录出现过的字符串），将其入队。 + +注意：一旦res不为空，就说明上一轮循环已经找到了合法的串，再进行循环的话会继续去掉更多括号，所以没必要继续循环。 + +本方法存在大量重复计算，所以比较耗时。 + + +## 思路二、DFS +注意到要求去掉最少的字符，所以返回的所有字符串的长度都是固定的（而且分别删除多少左右括号也是固定的）。而且我们可以提前把这个长度算出来，这样就可以使用DFS了。 + +为此，我们首先先统计多余的括号的数量，用left表示多余的左括号，right表示多余的右括号，一次遍历就可算出left和right，例 +``` +0 1 2 3 4 5 6 7 +( ) ) ) ( ( ( ) + +i = 0, left = 1, right = 0 +i = 1, left = 0, right = 0 +i = 2, left = 0, right = 1 +i = 3, left = 0, right = 2 +i = 4, left = 1, right = 2 +i = 5, left = 2, right = 2 +i = 6, left = 3, right = 2 +i = 7, left = 2, right = 2 +``` +上例有两个多余的左括号，两个多余的右括号。 + +然后我们就可以使用DFS： +* 当`left == 0 && right == 0`时，判断是否合法，若合法则存入结果数组res中； +* 否则，需要进行递归。遍历当前字符串，如果当前字符为左括号且`left > 0`，那么表示可以删掉这个左括号，然后递归调用；右括号同理。 + +注意为了避免重复计算，我们用变量start表示当前递归开始的位置，遍历时只需从start开始就可以了而不需要每次都从头开始。而且，对于相邻的相同括号，只需要删除第一个，比如 "())"，这里有两个右括号，不管删第一个还是删第二个右括号都会得到 "()"，没有区别，所以只用算一次就行了。 + +相比于思路一，本思路避免了很多重复计算，因此高效很多。 + +# C++ + +## 思路一 +``` C++ +class Solution { +private: + bool valid(const string &s){ // 判断字符串是否合法 + int cnt = 0; + for(int i = 0; i < s.size(); i++){ + if(s[i] == '(') cnt++; + else if(s[i] == ')' && (--cnt < 0)) return false; + } + return cnt == 0; + } + +public: + vector removeInvalidParentheses(string s) { + vectorres; + queueq; q.push(s); + unordered_setvisited; visited.insert(s); + + while(!q.empty()){ + string ss = q.front(); q.pop(); + if(valid(ss)) res.push_back(ss); + else{ + if(!res.empty()) continue; // 在上一轮已经找到了最长的合法串, 接下来的肯定要短些 + // 穷举:删除每个括号 + for(int i = 0; i < ss.size(); i++){ + if(ss[i] != '(' && ss[i] != ')') continue; + if(i > 0 && ss[i] == ss[i-1]) continue; + string subs = ss.substr(0, i) + ss.substr(i+1); + if(visited.find(subs) == visited.end()) { + q.push(subs); + visited.insert(subs); + } + } + } + } + + return res; + } +}; +``` + +## 思路二 + +``` C++ +class Solution { +private: + bool valid(const string &s){ + int cnt = 0; + for(int i = 0; i < s.size(); i++){ + if(s[i] == '(') cnt++; + else if(s[i] == ')' && (--cnt < 0)) return false; + } + return cnt == 0; + } + + void DFS(string s, int left, int right, int start, vector &res){ + if(!left && !right){ + if(valid(s)) res.push_back(s); + return; + } + for(int i = start; i < s.size(); i++){ + if(i > start && s[i] == s[i-1]) continue; // 去重 + + if(left > 0 && s[i] == '(') + DFS(s.substr(0, i) + s.substr(i+1), left-1, right, i, res); + else if(right > 0 && s[i] == ')') + DFS(s.substr(0, i) + s.substr(i+1), left, right-1, i, res); + } + } + +public: + vector removeInvalidParentheses(string s) { + vector res; + + int left = 0, right = 0; + for(int i = 0; i < s.size(); i++){ + if(s[i] == '(') left++; + else if(s[i] == ')'){ + if(left > 0) left--; + else right++; + } + } + + DFS(s, left, right, 0, res); + return res; + } +}; +``` \ No newline at end of file