[leetcode] Count of Smaller Numbers After Self
题目： 分析：很显然不能用时间复杂度为O(N2)的暴力穷举法，下面先给出一个运行时间为102ms的AC答案： 上面的答案虽然AC了，但若原始海量数据本身是降序排列时，再 ...
原文：leetcode 1365. How Many Numbers Are Smaller Than the Current Number
Given the arraynums, for eachnums i find out how many numbers in the array are smaller than it. That is, for eachnums i you have to count the number of validj ssuch thatj iandnums j lt nums i . Return ...
2020-03-05 22:13 0 929 推荐指数：
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