Leetcode 436. Find Right Interval 找区间 解题报告

1 解题思想

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的结束为止

其实暴力一点可以直接搜索，但是这里还不需要

这里使用了Java中的TreeMap
首先将所有起始位置和他的序号放入TreeMap（key是位置I的起始位置，value是I）当中

随后遍历每个位置的结束为止，使用TreeMap的方法，使用当前序号结束位置的大小找到TreeMap中第一个大于等于其结束位置的Entry，如果存在则取出value，不然就返回-1

2 原题

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:
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.

3 AC解

/**
 * 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[] result = new int[intervals.length];
        java.util.NavigableMap 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) {
            //使用treemap返回至少大于等于的给定值，也就是至少大于等于等于intervals[i].end的值
            Map.Entry entry = map.ceilingEntry(intervals[i].end);
            result[i] = (entry != null) ? entry.getValue() : -1;
        }

        return result;
    }
}