[LeetCode]Find Right Interval

题目描述:

LeetCode 436. Find Right Interval

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:

  1. You may assume the interval's end point is always bigger than its start point.
  2. 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:

给一组区间,对于每一个区间i,检查是否存在区间j,满足j的起点大于等于i的终点,我们称j在i的“右边"。

对于任意区间i,你需要存储j的最小下标,这意味着区间j拥有最小的起点并且位于i的“右边”。如果j不存在,则存储为-1。将最终结果以数组形式返回。

注意:

  1. 你可以假设区间的终点总是比起点大
  2. 你可以假设任意区间的起点都不相同

解题思路:

排序(Sort)+ 二分查找(Binary Search)

按照区间起点排序,然后二分查找即可。

Python代码:

# 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]
        """
        invs = sorted((x.start, i) for i, x in enumerate(intervals))
        ans = []
        for x in intervals:
            idx = bisect.bisect_right( invs, (x.end,) )
            ans.append(invs[idx][1] if idx < len(intervals) else -1)
        return ans

 

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