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

## # 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.
``````

## # 解题方法

``````# 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

``````

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