标签：bis ted poi after ble [1] nbsp point blog
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:
def binarySearchStartPoint(targetLst, ele): if len(targetLst) == 1: #only one element if targetLst[0][0].start >= ele: return targetLst[0][1] else: return -1 l = 0 h = len(targetLst)-1 while (l <= h): mid = (l + h)/2 if ele == targetLst[mid][0].start: return targetLst[mid][1] #index elif ele < targetLst[mid][0].start: h = mid - 1 else: l = mid + 1 #print ( ‘ddd : ‘, len(targetLst), l) if l >= len(targetLst): return -1 return targetLst[l][1] intersIndexLst = [(intl, i) for i, intl in enumerate(intervals)] intersIndexSortedLst = sorted(intersIndexLst, key = lambda k: k[0].start) i = 0 ansLst = [0] * len(intersIndexSortedLst) while (i < len(intersIndexSortedLst)-1): endElement = intersIndexSortedLst[i][0].end targetLst = intersIndexSortedLst[i+1:] currInd = intersIndexSortedLst[i][1] indRight = binarySearchStartPoint(targetLst, endElement) #print (‘targetLst: ‘, endElement, indRight) ansLst[currInd] = indRight i += 1 ansLst[intersIndexSortedLst[-1][1]] = -1 #the last emelement in intersIndexSortedLst return ansLst
2.
I refer to other‘s solution, which makes the question so simple.
(1) only need to maintain the start point with a tuple
(2) bisect can be used in that way,
(3) after sorted, directly compare the end with the original interval list to find the insertion position (index)
ansLst = [] intersIndexLst = [(intl.start, i) for i, intl in enumerate(intervals)] intersIndexSortedLst = sorted(intersIndexLst) for intl in intervals: ind = bisect_left(intersIndexSortedLst, (intl.end, )) ansLst.append(intersIndexSortedLst[ind][1] if ind < len(intervals) else - 1) return ansLst
--reference:
https://discuss.leetcode.com/topic/65596/python-o-nlogn-short-solution-with-explanation
[Leetcode] Binary search--436. Find Right Interval
标签：bis ted poi after ble [1] nbsp point blog
原文地址：http://www.cnblogs.com/anxin6699/p/6943178.html
