[leetcode] Count of Smaller Numbers After Self
題目： 分析：很顯然不能用時間復雜度為O(N2)的暴力窮舉法，下面先給出一個運行時間為102ms的AC答案： 上面的答案雖然AC了，但若原始海量數據本身是降序排列時，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 ...
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