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

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