README_EN.md

1190. Reverse Substrings Between Each Pair of Parentheses

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Description

You are given a string s that consists of lower case English letters and brackets.

Reverse the strings in each pair of matching parentheses, starting from the innermost one.

Your result should not contain any brackets.

 

Example 1:

Input: s = "(abcd)"
Output: "dcba"

Example 2:

Input: s = "(u(love)i)"
Output: "iloveu"
Explanation: The substring "love" is reversed first, then the whole string is reversed.

Example 3:

Input: s = "(ed(et(oc))el)"
Output: "leetcode"
Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.

 

Constraints:

• 1 <= s.length <= 2000
• s only contains lower case English characters and parentheses.
• It is guaranteed that all parentheses are balanced.

Solutions

Use deque or stack to simulate the reversal process.

Python3

class Solution:
    def reverseParentheses(self, s: str) -> str:
        stack = []
        for c in s:
            if c == ")":
                tmp = []
                while stack[-1] != "(":
                    tmp += stack.pop()
                stack.pop()
                stack += tmp
            else:
                stack.append(c)
        return "".join(stack)

Java

class Solution {
    public String reverseParentheses(String s) {
        Deque<Character> deque = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c == ')') {
                StringBuilder sb = new StringBuilder();
                while (deque.peekLast() != '(') {
                    sb.append(deque.pollLast());
                }
                deque.pollLast();
                for (int i = 0, n = sb.length(); i < n; i++) {
                    deque.offerLast(sb.charAt(i));
                }
            } else {
                deque.offerLast(c);
            }
        }
        StringBuilder sb = new StringBuilder();
        while (!deque.isEmpty()) {
            sb.append(deque.pollFirst());
        }
        return sb.toString();
    }
}

JavaScript

/**
 * @param {string} s
 * @return {string}
 */
var reverseParentheses = function (s) {
    let stack = [];
    let hashMap = {};
    const n = s.length;
    for (let i = 0; i < n; i++) {
        let cur = s.charAt(i);
        if (cur == '(') {
            stack.push(i);
        } else if (cur == ')') {
            let left = stack.pop();
            hashMap[left] = i;
            hashMap[i] = left;
        }
    }
    let res = [];
    let i = 0;
    let step = 1; // 1向右，-1向左
    while (i > -1 && i < n) {
        let cur = s.charAt(i);
        if (cur == '(' || cur == ')') {
            step = -step;
            i = hashMap[i];
        } else {
            res.push(cur);
        }
        i += step;
    }
    return res.join('');
};

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