1190. Reverse Substrings Between Each Pair of Parentheses
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
Solution 1: Simulation
We can directly use a stack to simulate the reversal process.
The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the length of the string $s$.
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Solution 2: Brain Teaser
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 $d$ to record the position of the corresponding other bracket for each (
or )
, i.e., $d[i]$ represents the position of the other bracket corresponding to the bracket at position $i$. We can directly use a stack to compute the array $d$.
Then, we traverse the string from left to right. When encountering (
or )
, we jump to the corresponding position according to the array $d$, then reverse the direction and continue traversing until the entire string is traversed.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string $s$.
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