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436 find right interval

admin 1月 08, 2021 0

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:

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Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

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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:

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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.

思路

利用哈希表存储临时的访问数据并不能防止TLE。这种方法不好使。

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class (object):
    def findRightInterval(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[int]
        """
        stats = {}
        length = len(intervals)
        ans = [-1] * length
        
        for i in range(length):
            cur = intervals[i].end
            if (stats.has_key(cur)):
                ans[i] = stats[cur]
            else:
                target = float('inf')
                pos = -1
                for j in range(length):
                    if (cur <= intervals[j].start):
                        if (target >= intervals[j].start):
                            target = intervals[j].start
                            pos = j
                ans[i] = pos
                stats[cur] = pos
        return ans

还是利用二分法的查找来解决问题。思路其实很简单，把所有的intervals.start作为一个数组，intervals.end作为一个数组。对start数组排序，根据end数组来找和start数组里面最接近的数字。

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# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e
class (object):
    def findRightInterval(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[int]
        """
        stats = {}
        find = {}
        length = len(intervals)
        ans = [-1] * length
        
        starts = []
        
        for i in range(length):
            find[intervals[i].start] = i
            starts.append(intervals[i].start)
        
        starts.sort()
        
        for i in range(length):
            cur = intervals[i].end
            if (stats.has_key(cur)):
                ans[i] = stats[cur]
            else:
                low = 0
                high = length - 1
                mid = -1
                useMid = False
                if (cur <= starts[-1]):
                    while (low + 1 < high):
                        mid = ( low + high ) / 2
                        if (starts[mid] < cur):
                            low = mid
                        elif(starts[mid] > cur):
                            high = mid
                        else:
                            useMid = True
                            break
                    if(useMid):
                        stats[cur] = find[starts[mid]]
                        ans[i] = find[starts[mid]]
                    else:
                        lowcur = starts[low] - cur
                        if (lowcur > 0):
                            stats[cur] = find[starts[low]]
                            ans[i] = find[starts[low]]
                        else:
                            stats[cur] = find[starts[high]]
                            ans[i] = find[starts[high]]
        return ans
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