# Find Right Interval

<https://leetcode.com/problems/find-right-interval/description/>

> 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:**
>
> 1. You may assume the interval's end point is always bigger than its start point.
> 2. You may assume none of these intervals have the same start point.

## Thoughts

和heaters异曲同工。只不过这次sort的目标换成了interval, 需要自己实现下Comparator.\
然后遍历intervals查找排好序的离该interval.end比它的最近的start是哪个。注意由于对返回的顺序有要求，因此要备份下位置。

## Code

```
/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    private int binSearch(Interval[] intervals, int target) {
        int start = 0, end = intervals.length - 1;
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (intervals[mid].start < target) {
                start = mid + 1;
            } else {
                end = mid;
            }
        }

        if (intervals[start].start >= target) {
            return start;
        }
        return -1;
    }


    public int[] findRightInterval(Interval[] intervals) {
        Map<Interval, Integer> locs = new HashMap<>();
        Interval[] tmp = new Interval[intervals.length];
        for (int i = 0; i < intervals.length; i++) {
            locs.put(intervals[i], i);
            tmp[i] = intervals[i];
        }
        Arrays.sort(tmp, (a, b) -> a.start == b.start ? a.end - b.end : a.start - b.start);
        int[] res = new int[intervals.length];
        for (int i = 0; i < intervals.length; i++) {
            int tpos = binSearch(tmp, intervals[i].end);
            if (tpos != -1) {
                res[i] = locs.get(tmp[tpos]);
            } else {
                res[i] = -1;
            }
        }
        return res;
    }
}
```

## Analysis

时间复杂度为O(nlgn).


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