436. Find Right Interval #
题目 #
Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the “right” of i.
For any interval i, you need to store the minimum interval j’s index, which means that the interval j has the minimum start point to build the “right” relationship for interval i. If the interval j doesn’t exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
Note:
- You may assume the interval’s end point is always bigger than its start point.
- You may assume none of these intervals have the same start point.
Example 1:
Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: [ [3,4], [2,3], [1,2] ]
Output: [-1, 0, 1]
Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.
Example 3:
Input: [ [1,4], [2,3], [3,4] ]
Output: [-1, 2, -1]
Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.
Note: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
题目大意 #
给定一组区间，对于每一个区间 i，检查是否存在一个区间 j，它的起始点大于或等于区间 i 的终点，这可以称为 j 在 i 的“右侧”。
对于任何区间，你需要存储的满足条件的区间 j 的最小索引，这意味着区间 j 有最小的起始点可以使其成为“右侧”区间。如果区间 j 不存在，则将区间 i 存储为 -1。最后，你需要输出一个值为存储的区间值的数组。
注意:
- 你可以假设区间的终点总是大于它的起始点。
- 你可以假定这些区间都不具有相同的起始点。
解题思路 #
- 给出一个
interval的 数组，要求找到每个interval在它右边第一个interval的下标。A 区间在 B 区间的右边：A 区间的左边界的值大于等于 B 区间的右边界。 - 这一题很明显可以用二分搜索来解答。先将
interval数组排序，然后针对每个interval，用二分搜索搜索大于等于interval右边界值的interval。如果找到就把下标存入最终数组中，如果没有找到，把-1存入最终数组中。
代码 #
package leetcode
import "sort"
// 解法一 利用系统函数 sort + 二分搜索
func findRightInterval(intervals [][]int) []int {
intervalList := make(intervalList, len(intervals))
// 转换成 interval 类型
for i, v := range intervals {
intervalList[i] = interval{interval: v, index: i}
}
sort.Sort(intervalList)
out := make([]int, len(intervalList))
for i := 0; i < len(intervalList); i++ {
index := sort.Search(len(intervalList), func(p int) bool { return intervalList[p].interval[0] >= intervalList[i].interval[1] })
if index == len(intervalList) {
out[intervalList[i].index] = -1
} else {
out[intervalList[i].index] = intervalList[index].index
}
}
return out
}
type interval struct {
interval []int
index int
}
type intervalList []interval
func (in intervalList) Len() int { return len(in) }
func (in intervalList) Less(i, j int) bool {
return in[i].interval[0] <= in[j].interval[0]
}
func (in intervalList) Swap(i, j int) { in[i], in[j] = in[j], in[i] }
// 解法二 手撸 sort + 二分搜索
func findRightInterval1(intervals [][]int) []int {
if len(intervals) == 0 {
return []int{}
}
intervalsList, res, intervalMap := []Interval{}, []int{}, map[Interval]int{}
for k, v := range intervals {
intervalsList = append(intervalsList, Interval{Start: v[0], End: v[1]})
intervalMap[Interval{Start: v[0], End: v[1]}] = k
}
quickSort(intervalsList, 0, len(intervalsList)-1)
for _, v := range intervals {
tmp := searchFirstGreaterInterval(intervalsList, v[1])
if tmp > 0 {
tmp = intervalMap[intervalsList[tmp]]
}
res = append(res, tmp)
}
return res
}
func searchFirstGreaterInterval(nums []Interval, target int) int {
low, high := 0, len(nums)-1
for low <= high {
mid := low + ((high - low) >> 1)
if nums[mid].Start >= target {
if (mid == 0) || (nums[mid-1].Start < target) { // 找到第一个大于等于 target 的元素
return mid
}
high = mid - 1
} else {
low = mid + 1
}
}
return -1
}