摘要：
My PriorityQueue Solution: class Solution { public int[][] merge(int[][] intervals) { PriorityQueue<int[]> queue = new PriorityQueue<>((a,b)->a[0]-b[0 阅读全文
posted @ 2022-04-07 06:33
阳光明媚的菲越
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摘要：
Just use BFS to solve this problem: 1. put the s to queue 2. if s is not a valid string, then remove a '(' or ')', and then put to the queue. 3. once 阅读全文
posted @ 2022-04-07 02:42
阳光明媚的菲越
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