python leetcode 436. Find Right Interval
一道很巧妙的題 假設第i個有右區間 那麼intervals[i].end<intervals[i+1].start
因為start唯一 所以我們建立一個列表裡面存的是(intervals[i].start,i) 再利用二分查詢尋找end位於列表中的哪個位置
擴充套件：假設end唯一 尋找左區間
class Solution: 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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