| comments | difficulty | edit_url | rating | source | tags | ||
|---|---|---|---|---|---|---|---|
true |
Medium |
1485 |
Weekly Contest 154 Q2 |
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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 <= 2000sonly contains lower case English characters and parentheses.- It is guaranteed that all parentheses are balanced.
We can directly use a stack to simulate the reversal process.
The time complexity is
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 {
public String reverseParentheses(String s) {
StringBuilder stk = new StringBuilder();
for (char c : s.toCharArray()) {
if (c == ')') {
StringBuilder t = new StringBuilder();
while (stk.charAt(stk.length() - 1) != '(') {
t.append(stk.charAt(stk.length() - 1));
stk.deleteCharAt(stk.length() - 1);
}
stk.deleteCharAt(stk.length() - 1);
stk.append(t);
} else {
stk.append(c);
}
}
return stk.toString();
}
}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;
}
};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)
}function reverseParentheses(s: string): string {
const stk: string[] = [];
for (const c of s) {
if (c === ')') {
const t: string[] = [];
while (stk.at(-1)! !== '(') {
t.push(stk.pop()!);
}
stk.pop();
stk.push(...t);
} else {
stk.push(c);
}
}
return stk.join('');
}/**
* @param {string} s
* @return {string}
*/
var reverseParentheses = function (s) {
const stk = [];
for (const c of s) {
if (c === ')') {
const t = [];
while (stk.at(-1) !== '(') {
t.push(stk.pop());
}
stk.pop();
stk.push(...t);
} else {
stk.push(c);
}
}
return stk.join('');
};We observe that, when traversing the string, each time we encounter ( or ), we jump to the corresponding ) or ( and then reverse the direction of traversal to continue.
Therefore, we can use an array ( or ), i.e.,
Then, we traverse the string from left to right. When encountering ( or ), we jump to the corresponding position according to the array
The time complexity is
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)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();
}
}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;
}
};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)
}function reverseParentheses(s: string): string {
const n = s.length;
const d: number[] = Array(n).fill(0);
const stk: number[] = [];
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: string[] = [];
while (i < n) {
const c = s.charAt(i);
if ('()'.includes(c)) {
i = d[i];
x = -x;
} else {
ans.push(c);
}
i += x;
}
return ans.join('');
}/**
* @param {string} s
* @return {string}
*/
var reverseParentheses = function (s) {
const n = s.length;
const d = 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 ('()'.includes(c)) {
i = d[i];
x = -x;
} else {
ans.push(c);
}
i += x;
}
return ans.join('');
};