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:
        stk = []
        for c in s:
            if c == ')':
                t = []
                while stk[-1] != '(':
                    t.append(stk.pop())
                stk.pop()
                stk.extend(t)
            else:
                stk.append(c)
        return ''.join(stk)

class Solution:
    def reverseParentheses(self, s: str) -> str:
        n = len(s)
        d = [0] * n
        stk = []
        for i, c in enumerate(s):
            if c == '(':
                stk.append(i)
            elif c == ')':
                j = stk.pop()
                d[i], d[j] = j, i
        i, x = 0, 1
        ans = []
        while i < n:
            if s[i] in '()':
                i = d[i]
                x = -x
            else:
                ans.append(s[i])
            i += x
        return ''.join(ans)

Java

class Solution {
    public String reverseParentheses(String s) {
        int n = s.length();
        int[] d = new int[n];
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '(') {
                stk.push(i);
            } else if (s.charAt(i) == ')') {
                int j = stk.pop();
                d[i] = j;
                d[j] = i;
            }
        }
        StringBuilder ans = new StringBuilder();
        int i = 0, x = 1;
        while (i < n) {
            if (s.charAt(i) == '(' || s.charAt(i) == ')') {
                i = d[i];
                x = -x;
            } else {
                ans.append(s.charAt(i));
            }
            i += x;
        }
        return ans.toString();
    }
}

C++

class Solution {
public:
    string reverseParentheses(string s) {
        string stk;
        for (char& c : s) {
            if (c == ')') {
                string t;
                while (stk.back() != '(') {
                    t.push_back(stk.back());
                    stk.pop_back();
                }
                stk.pop_back();
                stk += t;
            } else {
                stk.push_back(c);
            }
        }
        return stk;
    }
};

class Solution {
public:
    string reverseParentheses(string s) {
        int n = s.size();
        vector<int> d(n);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            if (s[i] == '(') {
                stk.push(i);
            } else if (s[i] == ')') {
                int j = stk.top();
                stk.pop();
                d[i] = j;
                d[j] = i;
            }
        }
        int i = 0, x = 1;
        string ans;
        while (i < n) {
            if (s[i] == '(' || s[i] == ')') {
                i = d[i];
                x = -x;
            } else {
                ans.push_back(s[i]);
            }
            i += x;
        }
        return ans;
    }
};

Go

func reverseParentheses(s string) string {
	stk := []byte{}
	for i := range s {
		if s[i] == ')' {
			t := []byte{}
			for stk[len(stk)-1] != '(' {
				t = append(t, stk[len(stk)-1])
				stk = stk[:len(stk)-1]
			}
			stk = stk[:len(stk)-1]
			stk = append(stk, t...)
		} else {
			stk = append(stk, s[i])
		}
	}
	return string(stk)
}

func reverseParentheses(s string) string {
	n := len(s)
	d := make([]int, n)
	stk := []int{}
	for i, c := range s {
		if c == '(' {
			stk = append(stk, i)
		} else if c == ')' {
			j := stk[len(stk)-1]
			stk = stk[:len(stk)-1]
			d[i], d[j] = j, i
		}
	}
	ans := []byte{}
	i, x := 0, 1
	for i < n {
		if s[i] == '(' || s[i] == ')' {
			i = d[i]
			x = -x
		} else {
			ans = append(ans, s[i])
		}
		i += x
	}
	return string(ans)
}

JavaScript

/**
 * @param {string} s
 * @return {string}
 */
var reverseParentheses = function (s) {
    const n = s.length;
    const d = new Array(n).fill(0);
    const stk = [];
    for (let i = 0; i < n; ++i) {
        if (s[i] == '(') {
            stk.push(i);
        } else if (s[i] == ')') {
            const j = stk.pop();
            d[i] = j;
            d[j] = i;
        }
    }
    let i = 0;
    let x = 1;
    const ans = [];
    while (i < n) {
        const c = s.charAt(i);
        if (c == '(' || c == ')') {
            i = d[i];
            x = -x;
        } else {
            ans.push(c);
        }
        i += x;
    }
    return ans.join('');
};

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