[LeetCode](week5) 315. Count of Smaller Numbers After Self

315. Count of Smaller Numbers After Self

题目

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Input: [5,2,6,1]
Output: [2,1,1,0] 
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

题目解析与作答

这个题目很简短，而且很清楚，可以先用最简单的穷举法写出一个解法：

V1 穷举法

class Solution {
public:
    vector countSmaller(vector& nums) {
        vector result;
        for(int i = 0; i < nums.size(); ++i){
            int count = 0;
            for(int e = i+1; e < nums.size(); ++e){
                if(nums[e] < nums[i]) 
                    count++;
            }
            result.push_back(count);
        }
        return result;
    }
};

这个解法竟然可以被Leetcode通过，实在是诧异。但也不妨来分析一下这个穷举法的复杂度：很显然，两层循环