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| 1 | +# 56. 合并区间 |
| 2 | + |
| 3 | +[url](https://leetcode-cn.com/problems/merge-intervals/) |
| 4 | + |
| 5 | +## 题目 |
| 6 | + |
| 7 | +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。 |
| 8 | + |
| 9 | +``` |
| 10 | +输入:intervals = [[1,3],[2,6],[8,10],[15,18]] |
| 11 | +输出:[[1,6],[8,10],[15,18]] |
| 12 | +解释:区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]. |
| 13 | +输入:intervals = [[1,4],[4,5]] |
| 14 | +输出:[[1,5]] |
| 15 | +解释:区间 [1,4] 和 [4,5] 可被视为重叠区间。 |
| 16 | +``` |
| 17 | + |
| 18 | +## 方法 |
| 19 | + |
| 20 | +## code |
| 21 | + |
| 22 | +### js |
| 23 | + |
| 24 | +```js |
| 25 | +let merge = intervals => { |
| 26 | + if (intervals === null || intervals.length <= 1) return intervals; |
| 27 | + intervals.sort((a, b) => { |
| 28 | + return a[0] - b[0] |
| 29 | + }); |
| 30 | + let list = [] |
| 31 | + let i = 0, n = intervals.length; |
| 32 | + while (i < n) { |
| 33 | + let l = intervals[i][0]; |
| 34 | + let r = intervals[i][1]; |
| 35 | + while (i < n - 1 && r >= intervals[i + 1][0]) { |
| 36 | + r = Math.max(r, intervals[i + 1][1]) |
| 37 | + i++; |
| 38 | + } |
| 39 | + list.push([l, r]); |
| 40 | + i++; |
| 41 | + } |
| 42 | + return list; |
| 43 | +} |
| 44 | +``` |
| 45 | + |
| 46 | +### go |
| 47 | + |
| 48 | +```go |
| 49 | +type numsSortable [][]int |
| 50 | + |
| 51 | +func (nums numsSortable) Len() int { |
| 52 | + return len(nums) |
| 53 | +} |
| 54 | +func (nums numsSortable) Less(i, j int) bool { |
| 55 | + return nums[i][0] <= nums[j][0] |
| 56 | +} |
| 57 | +func (nums numsSortable) Swap(i, j int) { |
| 58 | + nums[i], nums[j] = nums[j], nums[i] |
| 59 | +} |
| 60 | +func merge2(intervals [][]int) [][]int { |
| 61 | + Max := func(a, b int) int { |
| 62 | + if a > b { |
| 63 | + return a |
| 64 | + } |
| 65 | + return b |
| 66 | + } |
| 67 | + if len(intervals) <= 1{ |
| 68 | + return intervals |
| 69 | + } |
| 70 | + sort.Sort(numsSortable(intervals)) |
| 71 | + list := make([][]int, 0) |
| 72 | + i, n := 0, len(intervals) |
| 73 | + for i < n { |
| 74 | + l, r := intervals[i][0], intervals[i][1] |
| 75 | + for i < n-1 && r >= intervals[i+1][0] { |
| 76 | + r = Max(r, intervals[i + 1][1]) |
| 77 | + i++ |
| 78 | + } |
| 79 | + list = append(list, []int{l, r}) |
| 80 | + i++ |
| 81 | + } |
| 82 | + return list |
| 83 | +} |
| 84 | +``` |
| 85 | + |
| 86 | + |
| 87 | +### java |
| 88 | + |
| 89 | +```java |
| 90 | +class Solution { |
| 91 | + public int[][] merge(int[][] intervals) { |
| 92 | + if (intervals == null || intervals.length <= 1) return intervals; |
| 93 | + Arrays.sort(intervals, (a, b) -> a[0] - b[0]); |
| 94 | + List<int[]> list = new ArrayList<>(); |
| 95 | + int i = 0; |
| 96 | + int n = intervals.length; |
| 97 | + while (i < n) { |
| 98 | + int l = intervals[i][0]; |
| 99 | + int r = intervals[i][1]; |
| 100 | + while (i < n - 1 && r >= intervals[i + 1][0]) { |
| 101 | + r = Math.max(r, intervals[i + 1][1]); |
| 102 | + i++; |
| 103 | + } |
| 104 | + list.add(new int[] {l, r}); |
| 105 | + i++; |
| 106 | + } |
| 107 | + return list.toArray(new int[list.size()][2]); |
| 108 | + } |
| 109 | +} |
| 110 | +``` |
| 111 | + |
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