Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].Example 2:
Input: [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
合并区间。
题意是给一个二维数组，其中每个元素给的是一个类似[start, end]的范围。要求把这些区间尽可能地merge在一起。这题又是用到扫描线的思想。首先需要根据start对input排序，然后从第二个interval SECOND开始，跟其之前的interval FIRST作比较。比较的方式是
- 如果SECOND[start] > FIRST[end] - 说明两个interval没有交集，直接把FIRST加入最后的结果集
- 如果SECOND[start] <= FIRST[end] - 说明两个interval有交集，需要merge。merge后的interval的start自然是FIRST[start]，但是merge后的end是比较FIRST[end]和SECOND[end]谁大
时间O(nlogn) - 因为有sort
空间O(n) - 存放了最后的输出
JavaScript实现
1 /** 2 * @param {number[][]} intervals 3 * @return {number[][]} 4 */ 5 var merge = function (intervals) { 6 // corner case 7 if (intervals === null || intervals.length === 0) { 8 return intervals; 9 } 10 11 // normal case 12 let res = []; 13 intervals.sort((a, b) => a[0] - b[0]); 14 let cur = intervals[0]; 15 res.push(cur); 16 for (let interval of intervals) { 17 if (interval[0] <= cur[1]) { 18 cur[1] = Math.max(cur[1], interval[1]); 19 } else { 20 cur = interval; 21 res.push(cur); 22 } 23 } 24 return res; 25 };
Java实现
1 class Solution { 2 public int[][] merge(int[][] intervals) { 3 // corner case 4 if (intervals.length <= 1) { 5 return intervals; 6 } 7 8 // normal case 9 Arrays.sort(intervals, (a, b) -> a[0] - b[0]); 10 List<int[]> res = new ArrayList<>(); 11 int[] cur = intervals[0]; 12 res.add(cur); 13 for (int[] interval : intervals) { 14 if (interval[0] <= cur[1]) { 15 cur[1] = Math.max(cur[1], interval[1]); 16 } else { 17 cur = interval; 18 res.add(cur); 19 } 20 } 21 return res.toArray(new int[res.size()][]); 22 } 23 }