以数组 intervals 表示若干个区间的集合，其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间，并返回一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间。
示例 1：
输入：intervals = [[1,3],[2,6],[8,10],[15,18]]
输出：[[1,6],[8,10],[15,18]]
解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
class Solution {
public int[][] merge(int[][] intervals) {
if (intervals == null || intervals.length == 1) {
return intervals;
}
PriorityQueue<Node> minQueue = new PriorityQueue<>((a, b) -> a.start - b.start);
for (int i = 0; i < intervals.length; i++) {
minQueue.add(new Node(intervals[i][0], intervals[i][1]));
}
int[] flag = new int[2];
Node curNode = minQueue.poll();
flag[0] = curNode.start;
flag[1] = curNode.end;
int leftMin = flag[0];
int rightMax = flag[1];
PriorityQueue<Node> queue = new PriorityQueue<>((a, b) -> a.start - b.start);
while (!minQueue.isEmpty()) {
Node cur = minQueue.poll();
if (flag[1] >= cur.start) {
leftMin = Math.min(leftMin, cur.start);
rightMax = Math.max(rightMax, cur.end);
flag[0] = leftMin;
flag[1] = rightMax;
} else if (cur.end <= rightMax) {
continue;
} else {
queue.add(new Node(leftMin, rightMax));
flag[0] = cur.start;
flag[1] = cur.end;
leftMin = flag[0];
rightMax = flag[1];
}
}
queue.add(new Node(leftMin, rightMax));
int[][] res = new int[queue.size()][2];
int index = 0;
while (!queue.isEmpty()) {
Node cur = queue.poll();
res[index][0] = cur.start;
res[index][1] = cur.end;
index++;
}
return res;
}
public class Node {
public int start;
public int end;
public Node(int s, int e) {
this.start = s;
this.end = e;
}
}
}
}