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.
解法: 把这些interval按照start从小到大排序，然后对每一个interval用其end去在排好序的队列里面做二分查找， 找到符合要求的一个interval。代码：
public int[] findRightInterval(Interval[] intervals){ Interval[] sortedIntervals = Arrays.copyOf(intervals,intervals.length); Arrays.sort(sortedIntervals,(o1, o2) -> o1.start - o2.start); int[] result = new int[intervals.length]; for (int i = 0; i < intervals.length; i++) { Interval current = intervals[i]; int insertIndex = Arrays.binarySearch(sortedIntervals, current, (o1, o2) -> o1.start - o2.end); if (insertIndex < 0){ insertIndex = -insertIndex - 1; } if (insertIndex == intervals.length){ result[i] = -1; }else { Interval match = sortedIntervals[insertIndex]; for (int j = 0; j < intervals.length; j++){ if (i != j && match.start == intervals[j].start && match.end == intervals[j].end){ // System.out.println(",old index:"+j); result[i] = j; } } } } return result; }