1021. Remove Outermost Parentheses

🟢   🔖  栈 字符串  🔗 LeetCode

题目

A valid parentheses string is either empty "", "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.

• For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string s is primitive if it is nonempty, and there does not exist a way to split it into s = A + B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string s, consider its primitive decomposition: s = P1 + P2 + ... + Pk, where Pi are primitive valid parentheses strings.

Return s after removing the outermost parentheses of every primitive string in the primitive decomposition ofs.

Example 1:

Input: s = "(()())(())"

Output: "()()()"

Explanation:

The input string is "(()())(())", with primitive decomposition "(()())" + "(())".

After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

Input: s = "(()())(())(()(()))"

Output: "()()()()(())"

Explanation:

The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".

After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Example 3:

Input: s = "()()"

Output: ""

Explanation:

The input string is "()()", with primitive decomposition "()" + "()".

After removing outer parentheses of each part, this is "" + "" = "".

Constraints:

• 1 <= s.length <= 10^5
• s[i] is either '(' or ')'.
• s is a valid parentheses string.

题目大意

题目要求去掉最外层的括号。

解题思路

用栈模拟，当遇到 ( 时入栈，当遇到 ) 时入栈，只有当栈内元素个数大于 1 时（去掉最外层的括号），才将当前字符累加到输出的字符串 str 上。

代码

/**
 * @param {string} s
 * @return {string}
 */
var removeOuterParentheses = function (s) {
	let stack = [];
	let str = '';
	for (let i = 0; i < s.length; i++) {
		if (s[i] === '(') {
			stack.push(s[i]);
			if (stack.length > 1) {
				str += s[i];
			}
		} else if (s[i] === ')') {
			if (stack.length > 1) {
				str += s[i];
			}
			stack.pop();
		}
	}
	return str;
};