436. Find Right Interval

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.

题目含义：

题目给了一堆[起始位置，结束位置]的数组，定义了一个个区间。

任务则是要求对于给定的第I个区间，找到一个最小的j，这里的j的起始位置大于等于I的结束位置。找不到为-1

    public int[] findRightInterval(Interval[] intervals) {
        int[] result = new int[intervals.length];
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for (int i = 0; i < intervals.length; i++) map.put(intervals[i].start, i);
        for (int i = 0; i < intervals.length; i++) {
            // lowerEntry、floorEntry、ceilingEntry 和 higherEntry 方法，它们分别返回与小于、小于等于、大于等于、大于给定键的键关联的 Map.Entry 对象。
            Map.Entry<Integer, Integer> item = map.ceilingEntry(intervals[i].end);//找出大于等于intervals[i].end的key
            result[i] = (item == null) ? -1 : item.getValue();
        }
        return result;        
    }