436. Find Right Interval

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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:

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Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

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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:

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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1：

这里遇到了一种binarysearch的用法，题目要求需要返回原始的index，而又有点像binarysearch
可以先用hashmap记录每一个当前的index，再对数组进行排序
排好序之后得到结果了再取map中找寻原始的index。
C++

1

Java

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/**
 * 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; }
 * }
 */
public class Solution {
    public int[] findRightInterval(Interval[] intervals) {
        HashMap<Integer, Integer> startMap = new HashMap<Integer, Integer>();
        List<Integer> starts = new ArrayList<Integer>();
        for (int i = 0; i < intervals.length; i++) {
            startMap.put(intervals[i].start, i);
            starts.add(intervals[i].start);
        }        
        Collections.sort(starts);   // sort the array of start
        int[] res = new int[starts.size()];
        for (int i = 0; i < intervals.length; i++) {
            int end = intervals[i].end;
            int temp = binarySearch(starts, end);
            if (temp == -1) {
                res[i] = -1;
            } else {
                res[i] = startMap.get(starts.get(temp));
            }
        }
        return res;        
    }
    private int binarySearch(List<Integer> list, int target) {
        // find first that is larger than end
        int start = 0, end = list.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (list.get(mid) < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (list.get(start) >= target) {
            return start;
        }
        if (list.get(end) >= target) {
            return end;
        }
        return -1;  // didn't find a start that is larger than target
    }
}