问题：
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 1:1
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5Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:1
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7Input: [ [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 3:1
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6Input: [ [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.
解答1
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27/*
* 道题给了我们一堆区间，让我们找每个区间的最近右区间，要保证右区间的start要大于等于当前区间的end，
* 由于区间的顺序不能变，所以我们不能给区间排序，我们需要建立区间的start和该区间位置之间的映射，
* 由于题目中限定了每个区间的start都不同，所以不用担心一对多的情况出现。然后我们把所有的区间的start
* 都放到一个数组中，并对这个数组进行降序排序，那么start值大的就在数组前面。然后我们遍历区间集合，对于
* 每个区间，我们在数组中找第一个小于当前区间的end值的位置，如果数组中第一个数就小于当前区间的end，那么说明该区间不存在右区间，
* 结果res中加入-1；如果找到了第一个小于当前区间end的位置，那么往前推一个就是第一个大于等于
* 当前区间end的start，我们在哈希表中找到该区间的坐标加入结果res中即可
*/
public class L436 {
/*
* 按照start排序，此题是用treemap来排序的，然后按照上面的方法找出第一个大于等于end的位置
*/
public int[] findRightInterval(int[][] intervals) {
int [] result = new int [intervals.length];
TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
for(int i = 0; i < intervals.length; i ++) {
map.put(intervals[i][0], i);
}
for(int i = 0; i < intervals.length; i ++) {
//使用treemap返回至少大于等于给定值，也就是至少大于等于intervals[i][1]的值
Map.Entry<Integer, Integer> entry = map.ceilingEntry(intervals[i][1]);
result[i] = (entry != null) ? entry.getValue() : -1;
}
return result;
}
}
