|
| 1 | +# 436. Find Right Interval |
| 2 | + |
| 3 | +**<font color=red>难度: Medium</font>** |
| 4 | + |
| 5 | +## 刷题内容 |
| 6 | + |
| 7 | +> 原题连接 |
| 8 | + |
| 9 | +* https://leetcode.com/problems/find-right-interval/description/ |
| 10 | + |
| 11 | +> 内容描述 |
| 12 | + |
| 13 | +``` |
| 14 | +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. |
| 15 | + |
| 16 | +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. |
| 17 | + |
| 18 | +Note: |
| 19 | +You may assume the interval's end point is always bigger than its start point. |
| 20 | +You may assume none of these intervals have the same start point. |
| 21 | +Example 1: |
| 22 | +Input: [ [1,2] ] |
| 23 | + |
| 24 | +Output: [-1] |
| 25 | + |
| 26 | +Explanation: There is only one interval in the collection, so it outputs -1. |
| 27 | +Example 2: |
| 28 | +Input: [ [3,4], [2,3], [1,2] ] |
| 29 | + |
| 30 | +Output: [-1, 0, 1] |
| 31 | + |
| 32 | +Explanation: There is no satisfied "right" interval for [3,4]. |
| 33 | +For [2,3], the interval [3,4] has minimum-"right" start point; |
| 34 | +For [1,2], the interval [2,3] has minimum-"right" start point. |
| 35 | +Example 3: |
| 36 | +Input: [ [1,4], [2,3], [3,4] ] |
| 37 | + |
| 38 | +Output: [-1, 2, -1] |
| 39 | + |
| 40 | +Explanation: There is no satisfied "right" interval for [1,4] and [3,4]. |
| 41 | +For [2,3], the interval [3,4] has minimum-"right" start point. |
| 42 | +``` |
| 43 | + |
| 44 | +## 解题方案 |
| 45 | + |
| 46 | +> 思路 1 |
| 47 | +******- 时间复杂度: O(N^2)******- 空间复杂度: O(N)****** |
| 48 | + |
| 49 | + |
| 50 | +打算先按照start,end排个序,然后只要往后面找就行,虽然是O(N^2),但是稍微小了点,结果超时。 |
| 51 | + |
| 52 | +``` |
| 53 | +class Solution(object): |
| 54 | + def findRightInterval(self, intervals): |
| 55 | + """ |
| 56 | + :type intervals: List[Interval] |
| 57 | + :rtype: List[int] |
| 58 | + """ |
| 59 | + res = [-1] * len(intervals) |
| 60 | + new_intervals = sorted(intervals, key = lambda x: (x.start, x.end)) |
| 61 | + lookup = {} |
| 62 | + for idx, interval in enumerate(intervals): |
| 63 | + lookup[(interval.start, interval.end)] = idx |
| 64 | + |
| 65 | + for i in range(len(new_intervals)): |
| 66 | + for j in range(i+1, len(new_intervals)): |
| 67 | + if new_intervals[j].start >= new_intervals[i].end: |
| 68 | + look_interval = (new_intervals[i].start, new_intervals[i].end) |
| 69 | + look_idx = lookup[look_interval] |
| 70 | + next_interval = (new_intervals[j].start, new_intervals[j].end) |
| 71 | + next_idx = lookup[next_interval] |
| 72 | + res[look_idx] = next_idx |
| 73 | + break |
| 74 | + return res |
| 75 | +``` |
| 76 | + |
| 77 | +> 思路 2 |
| 78 | +******- 时间复杂度: O(N^2)******- 空间复杂度: O(N)****** |
| 79 | + |
| 80 | +后面我想到既然所有的start都不一样,我不如搞一个list存排过序的start,然后这样就可以用二分查找来找到第一个大于某end的idx了, |
| 81 | +同样的用一个字典来存start和idx的信息 |
| 82 | + |
| 83 | +```python |
| 84 | +``` |
| 85 | + |
| 86 | + |
| 87 | + |
| 88 | + |
| 89 | + |
| 90 | + |
| 91 | +## Follow up |
| 92 | + |
| 93 | + |
| 94 | +就是题目变成 |
| 95 | +``` |
| 96 | +Given a set of intervals, for each of the interval i, |
| 97 | +check if there exists an interval j whose start point is bigger than or equal to the start point of the interval i, |
| 98 | +which can be called that j is on the "right" of i. |
| 99 | +``` |
| 100 | + |
| 101 | + |
| 102 | +> 思路 1 |
| 103 | +******- 时间复杂度: O(NlgN)******- 空间复杂度: O(N)****** |
| 104 | + |
| 105 | + |
| 106 | +```python |
| 107 | +class Solution(object): |
| 108 | + def findRightInterval(self, intervals): |
| 109 | + """ |
| 110 | + :type intervals: List[Interval] |
| 111 | + :rtype: List[int] |
| 112 | + """ |
| 113 | + res = [-1] * len(intervals) |
| 114 | + new_intervals = sorted(intervals, key = lambda x: (x.start, x.end)) |
| 115 | + lookup = {} |
| 116 | + for idx, interval in enumerate(intervals): |
| 117 | + lookup[(interval.start, interval.end)] = idx |
| 118 | + for i in range(len(new_intervals)-1): |
| 119 | + look_interval = (new_intervals[i].start, new_intervals[i].end) |
| 120 | + look_idx = lookup[look_interval] |
| 121 | + next_interval = (new_intervals[i+1].start, new_intervals[i+1].end) |
| 122 | + next_idx = lookup[next_interval] |
| 123 | + res[look_idx] = next_idx |
| 124 | + return res |
| 125 | +``` |
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