题目描述

• 重叠区间：需要有重叠判断
• 注意：题目并没有说集合间有序，因此要先做一个排序，以左下标为排序值（否则会出错
  

代码 & 思路

• 经过排序后，只需要循环一次即可，从左到右：可以合并则合并，不能合并则添加当前数组到答案中，然后开启新的数组继续。

class Solution {public int[][] merge(int[][] intervals) {// 区间个数int len = intervals.length;// 先进行排序O(nlogn)Arrays.sort(intervals,new Comparator<int[]>(){@Overridepublic int compare(int[] a,int[] b){return a[0]-b[0];}});List<int[]> ansL = new ArrayList<>();// 先加入一个初始数组int[] first = new int[2];first[0] = intervals[0][0];first[1] = intervals[0][1];ansL.add(first);int[] nowAL = ansL.get(0);// O(n)，一次遍历结束for(int i=1; i < len; i++){// 合并判断if(over(nowAL[0],nowAL[1],intervals[i][0],intervals[i][1])){// 更新区间值判断if(intervals[i][0]<nowAL[0]){nowAL[0] = intervals[i][0];}if(intervals[i][1]>nowAL[1]){nowAL[1] = intervals[i][1];}}// 不合并，则更新else{first = new int[2];first[0] = intervals[i][0];first[1] = intervals[i][1];ansL.add(first);nowAL = first;}}int[][] ans = new int[ansL.size()][2];for(int i=0;i<ansL.size();i++){ans[i] = ansL.get(i);}return ans;}// 重叠判断boolean over(int left1,int right1,int left2,int right2){// 不重叠的情况if(left1 > right2 || left2 > right1){return false;}return true;}
}

更新版 2.0

class Solution {public int[][] merge(int[][] intervals) {if(intervals == null || intervals.length <= 1) {return intervals;} List<int[]> list = new ArrayList<>();// 先排序 O(nlogN)Arrays.sort(intervals, (a, b) -> (a[0] - b[0]));// 时间复杂度 O(n)，每次合并第 i ~ j 个区间for(int i = 0, j; i < intervals.length; i = j) {// 定义当前左、右区间int left = intervals[i][0];int right = intervals[i][1];j = i + 1;for(; j < intervals.length && right >= intervals[j][0]; j++) {// 维护右区间right = Math.max(right, intervals[j][1]);}list.add(new int[]{left, right});}return list.toArray(new int[list.size()][2]);}
}

• 去注释最简版

class Solution {public int[][] merge(int[][] intervals) {if(intervals == null || intervals.length <= 1) {return intervals;} List<int[]> list = new ArrayList<>();Arrays.sort(intervals, (a, b) -> (a[0] - b[0]));for(int i = 0, j; i < intervals.length; i = j) {int left = intervals[i][0];int right = intervals[i][1];j = i + 1;for(; j < intervals.length && right >= intervals[j][0]; j++) {right = Math.max(right, intervals[j][1]);}list.add(new int[]{left, right});}return list.toArray(new int[list.size()][2]);}
}