436. Find Right Interval （寻找最近的右区间）

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.

题目大意：给定一些区间，找到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，返回值为最近右区间在数组中的下标，如果某个区间不存在右区间，即该区间的end比所有区间的最大的start还要大，则返回-1。

解题思路：先复制一份给定的区间数组，命名为sortedIntervals[]，在该副本中，修改区间的end属性值，用以记录该区间在原数组中的下标，然后对该副本数组sortedIntervals[]按照start值从小到大进行排序。然后遍历原数组intervals[]，对于其中的每一个区间对象，通过二分查找在排好序的sortedIntervals[]数组中查找最近的start值（要求在右边），然后返回对应的区间对象在原始数组中的下标值（现在存放在end属性中），对于end值比sortedIntervals[]中最大的start值还大的对象，返回-1。

代码如下（本以为效率很低，没想到提交之后发现22ms，beats 93.65%）

/** * 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) {int n = intervals.length;int[] res = new int[n];int index = 0;Interval[] sortedIntervals = new Interval[n];//复制原始数组，用end值存放该区间对象在原始数组中的下标值for(Interval interval : intervals){sortedIntervals[index] = new Interval(interval.start, index);index++;}List<Interval> list = new ArrayList<>(Arrays.asList(sortedIntervals));Collections.sort(list, new Comparator<Interval>() {@Overridepublic int compare(Interval o1, Interval o2) {return o1.start - o2.start;}});sortedIntervals = (Interval[]) list.toArray(new Interval[n]);int max = sortedIntervals[n - 1].start;index = 0;for (Interval interval : intervals) {if (interval.end > max)res[index] = -1;elseres[index] = binarySearch(sortedIntervals, interval.end);index++;}return res;}public int binarySearch(Interval[] intervals, int target) {int left = 0, right = intervals.length - 1;int mid;while (left <= right) {mid = (left + right) / 2;if (intervals[mid].start == target) {return intervals[mid].end;} else if (intervals[mid].start < target) {left = mid + 1;} else {right = mid - 1;}}return intervals[left].end;}}