436. Find Right Interval**

最新推荐文章于 2017-12-09 16:31:41 发布

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最新推荐文章于 2017-12-09 16:31:41 发布

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分类专栏： leetcode Java Binary Search

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本文介绍了一种算法，用于解决给定一系列区间时如何找到每个区间右侧最近且符合条件的另一个区间的索引问题。通过使用TreeMap，该算法能有效地找到对于每一个输入区间，其右侧开始位置最小的区间。

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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:

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.

    public int[] findRightInterval(Interval[] intervals) {
        TreeMap<Integer, Integer> map = new TreeMap<>();
        int[] res = new int[intervals.length];
        for(int i=0; i<intervals.length;i++){
            map.put(intervals[i].start, i);
        }
        for(int i=0;i<intervals.length;i++){
            Integer key = map.ceilingKey(intervals[i].end);
            res[i]=key!=null?map.get(key):-1;
        }
        return res;
    }

总结：好聪明，关键在于没有相同的start point。