436. Find Right Interval ——本质:查找题目,因此二分!
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.
# Definition for an interval. # class Interval(object): # def __init__(self, s=0, e=0): # self.start = s # self.end = e class Solution(object): def findRightInterval(self, intervals): """ :type intervals: List[Interval] :rtype: List[int] Input: [ [3,4], [2,3], [1,2] ] Output: [-1, 0, 1] [1, 2], [2, 3], [3, 4] 2, 1, 0 1, 0, -1 Input: [ [1,4], [2,3], [3,4] ] Output: [-1, 2, -1] [1, 4], [2, 3], [3, 4], [4, 5] sorted """ from bisect import bisect_left starts = [] pos_dict = {} for i,v in enumerate(intervals): pos_dict[v.start] = i starts.append(v.start) starts.sort() ans = [-1]*len(intervals) for i,v in enumerate(intervals): pos = bisect_left(starts, v.end) if pos>=0 and pos<len(intervals): ans[i] = pos_dict[starts[pos]] return ans