【LeetCode】436. Find Right Interval 解题报告（Python）

标签（空格分隔）： LeetCode

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目地址：https://leetcode.com/problems/find-right-interval/description/

题目描述：

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

给了一堆区间，找出每个区间右边最近的区间。不允许重合，每个区间的起始点不重复。如果不存在就返回-1.

解题方法

这个题主要是要使用二叉搜索，我发现我对这个理解的不够深入。

做法还是很容易理解的，因为可以使用一个字典保存每个区间的索引，因为每个区间的起点都是不同的，所以可以使用这个开始点当做区间的标记。

对起始点进行排序之后（为什么要排序？因为我们要使用二分查找），遍历每个区间，找出比这个区间的结尾大的第一个区间的起点值，然后根据这个起点值再找到这个区间的索引。

这也就是lowwer_found和higher_fount。我要补一补这方面的内容了。

代码如下：

# Definition for an interval.

# class Interval:

#     def __init__(self, s=0, e=0):

#         self.start = s

#         self.end = e

class Solution:

    def findRightInterval(self, intervals):

        """

        :type intervals: List[Interval]

        :rtype: List[int]

        """

        n = len(intervals)

        start_map = {interval.start : i for i, interval in enumerate(intervals)}

        start_list = [interval.start for interval in intervals]

        res = []

        start_list.sort()

        for interval in intervals:

            pos = self.higher_find(start_list, interval.end)

            res.append(start_map[start_list[pos]] if pos != -1 else -1)

        return res

    def higher_find(self, array, v):

        lo, hi = 0, len(array) - 1

        first = -1

        while lo <= hi:

            mid = lo + (hi - lo) // 2

            if array[mid] >= v:

                hi = mid - 1

                first = mid

            else:

                lo = mid + 1

        return first

参考资料：

https://leetcode.com/problems/find-right-interval/discuss/156832/Python-O(n*log(n))-O(n)-slow-AF

日期

2018 年 9 月 13 日 ———— 越刷越受挫