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