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.

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.

Solution

Binary search, time O(n * logn), space O(n)

按照start进行排序，并保留原始index。然后二分搜索first index equal or larger than target.

/**
 * 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 {
    public int[] findRightInterval(Interval[] intervals) {
        int n = intervals.length;
        List<int[]> startToIndex = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            int[] a = new int[2];
            a[0] = intervals[i].start;
            a[1] = i;
            startToIndex.add(a);
        }
        
        Collections.sort(startToIndex, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                return a[0] - b[0];
            } 
        });
        
        int[] res = new int[n];
        for (int i = 0; i < n; ++i) {
            res[i] = search(startToIndex, intervals[i].end);
        }
        
        return res;
    }
    
    public int search(List<int[]> nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (target > nums.get(mid)[0]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return (left < 0 || left >= nums.size()) ? -1 : nums.get(left)[1];
    }
}