Leetcode 436. Find Right Interval 找区间 解题报告
2016-11-06 14:33
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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<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) { //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<Integer, Integer> entry = map.ceilingEntry(intervals[i].end); result[i] = (entry != null) ? entry.getValue() : -1; } return result; } }
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