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.
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>() { @Override public 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; else res[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; } }
