这里解法是直接找一个快排模板，然后排序了第一个数，所以需要多加一个判断逻辑并且稍微覆盖。

package mainimport "fmt"func main() {fmt.Println(merge([][]int{{0, 2}, {2, 3}, {4, 4}, {0, 1}, {5, 7}, {4, 5}, {0, 0}}))// fmt.Println(merge([][]int{{2, 3}, {4, 5}, {6, 7}, {8, 9}, {1, 10}}))// fmt.Println(merge([][]int{{1, 4}, {1, 5}, {0, 0}}))
}func merge(intervals [][]int) [][]int {if len(intervals) == 0 {return intervals}quickSort(intervals, 0, len(intervals)-1)res := make([][]int, 0)res = append(res, intervals[0])tmpmax := 0 // 可能有包含关系for i := 1; i < len(intervals); i++ {//快排时没有处理第二位数字，会有1,5;1,4这种错误交换if intervals[i][0] == intervals[i-1][0] {tmpmax = max(intervals[i][1], intervals[i-1][1])// 并且暴力地把他们都用最大数覆盖了intervals[i][1], intervals[i-1][1] = tmpmax, tmpmaxres[len(res)-1][1] = max(res[len(res)-1][1], tmpmax)continue}if intervals[i][0] <= intervals[i-1][1] || res[len(res)-1][1] >= intervals[i][1] || res[len(res)-1][1] >= intervals[i][0] {// 已经加入res的可能对后面独立区间有包含关系tmpmax = max(intervals[i][1], intervals[i-1][1])res[len(res)-1][1] = max(res[len(res)-1][1], tmpmax)} else {res = append(res, intervals[i])}}return res
}func quickSort(nums [][]int, l, r int) { //[l,r]if l < r {m := partition(nums, l, r)quickSort(nums, l, m-1)quickSort(nums, m+1, r)}
}func partition(nums [][]int, l int, r int) int {key := nums[r][0]i := lj := lfor j < r {if nums[j][0] < key {nums[i][0], nums[j][0] = nums[j][0], nums[i][0]nums[i][1], nums[j][1] = nums[j][1], nums[i][1]i++}j++}nums[i][0], nums[r][0] = nums[r][0], nums[i][0]nums[i][1], nums[r][1] = nums[r][1], nums[i][1]return i
}

假如把第二个数一起排序了：

package mainimport "fmt"func main() {// fmt.Println(merge([][]int{{0, 2}, {2, 3}, {4, 4}, {0, 1}, {5, 7}, {4, 5}, {0, 0}}))// fmt.Println(merge([][]int{{2, 3}, {4, 5}, {6, 7}, {8, 9}, {1, 10}}))fmt.Println(merge([][]int{{1, 10}, {2, 3}, {4, 5}}))
}func merge(intervals [][]int) [][]int {if len(intervals) == 0 {return intervals}quickSort(intervals, 0, len(intervals)-1)res := make([][]int, 0)res = append(res, intervals[0])tmpmax := 0 // 可能有包含关系for i := 1; i < len(intervals); i++ {// 如果上一个区间完全包含于res，最后一个res的可能对后面的“看似独立区间”有包含关系// {1，10}，{2，3}，{4，5} // {1，10}，{2，3}，{4，15}if intervals[i][0] <= intervals[i-1][1] || res[len(res)-1][1] >= intervals[i][1] || res[len(res)-1][1] >= intervals[i][0] {tmpmax = max(intervals[i][1], intervals[i-1][1])res[len(res)-1][1] = max(res[len(res)-1][1], tmpmax)} else {res = append(res, intervals[i])}}return res
}func quickSort(nums [][]int, l, r int) { //[l,r]if l < r {m := partition(nums, l, r)quickSort(nums, l, m-1)quickSort(nums, m+1, r)}
}func partition(nums [][]int, l int, r int) int {key := nums[r][0]i := lj := lfor j < r {// 第一个数相同时，第二数也要小才是真的小if (nums[j][0] < key) || (nums[j][0] == key && nums[j][1] < nums[r][1]) {nums[i][0], nums[j][0] = nums[j][0], nums[i][0]nums[i][1], nums[j][1] = nums[j][1], nums[i][1]i++}j++}nums[i][0], nums[r][0] = nums[r][0], nums[i][0]nums[i][1], nums[r][1] = nums[r][1], nums[i][1]return i
}