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Commit 4c77c7a

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author
zhangxing
committed
feat: 栈 - 删除最外层的括号
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‎栈 - 删除最外层的括号.js‎

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// https://leetcode.cn/problems/remove-outermost-parentheses/
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// 有效括号字符串为空 ""、"(" + A + ")" 或 A + B ,其中 A 和 B 都是有效的括号字符串,+ 代表字符串的连接。
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// 例如,"","()","(())()" 和 "(()(()))" 都是有效的括号字符串。
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// 如果有效字符串 s 非空,且不存在将其拆分为 s = A + B 的方法,我们称其为原语(primitive),其中 A 和 B 都是非空有效括号字符串。
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// 给出一个非空有效字符串 s,考虑将其进行原语化分解,使得:s = P_1 + P_2 + ... + P_k,其中 P_i 是有效括号字符串原语。
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// 对 s 进行原语化分解,删除分解中每个原语字符串的最外层括号,返回 s 。
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//
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// 示例 1:
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// 输入:s = "(()())(())"
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// 输出:"()()()"
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// 解释:
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// 输入字符串为 "(()())(())",原语化分解得到 "(()())" + "(())",
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// 删除每个部分中的最外层括号后得到 "()()" + "()" = "()()()"。
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// 示例 2:
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// 输入:s = "(()())(())(()(()))"
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// 输出:"()()()()(())"
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// 解释:
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// 输入字符串为 "(()())(())(()(()))",原语化分解得到 "(()())" + "(())" + "(()(()))",
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// 删除每个部分中的最外层括号后得到 "()()" + "()" + "()(())" = "()()()()(())"。
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// 示例 3:
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// 输入:s = "()()"
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// 输出:""
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// 解释:
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// 输入字符串为 "()()",原语化分解得到 "()" + "()",
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// 删除每个部分中的最外层括号后得到 "" + "" = ""。
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var removeOuterParentheses = function (s) {
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let stack = [];
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let result = '';
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for (let i = 0; i < s.length; i++) {
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if (s[i] === '(') {
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stack.push(s[i]);
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}
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if (s[i] === ')') {
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stack.pop();
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}
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// 当栈为空时说明一个原语就结束了
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if (stack.length) {
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res += s[i];
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}
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}
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return result;
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};
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console.log(removeOuterParentheses('(()())(())'));
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console.log(removeOuterParentheses('(()())(())(()(()))'));
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