LeetCode题目：315. Count of Smaller Numbers After Self

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题目描述：You are given an integer array nums and you have to return a newcounts array. Thecounts array has the property where counts[i] is the number of smaller elements to the right ofnums[i].

Example:

Given nums = [5, 2, 6, 1]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.

Return the array [2, 1, 1, 0].

一开始直接就用暴力比较，但超时了。代码如下：

class Solution {public:    vector<int> countSmaller(vector<int>& nums) {        int l=nums.size();        if(l==0)return nums;        vector<int> c;        int num=0;        for(int i=0;i<l-1;i++){            for(int j=i+1;j<l;j++){                if(nums[i]>nums[j])num++;            }            c.push_back(num);            num=0;        }        c.push_back(0);        return c;    }}; 

在论坛中发现一个简单的代码，虽然accept了但是很多人认为O(n^2)的复杂度并不好：

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class Solution {public:    vector<int> countSmaller(vector<int>& nums) {        vector<int> db;        vector<int> result(nums.size());        for(int i = nums.size()-1; i >= 0; i--)        {            auto it = lower_bound(db.begin(), db.end(), nums[i]);            result[i] = it - db.begin();            db.insert(it, nums[i]);        }        return result;    }};

正确的代码如下：

class Solution {protected:    void merge_countSmaller(vector<int>& indices, int first, int last, vector<int>& results, vector<int>& nums) {        int count = last - first;        if (count > 1) {            int step = count / 2;            int mid = first + step;            merge_countSmaller(indices, first, mid, results, nums);            merge_countSmaller(indices, mid, last, results, nums);            vector<int> tmp;            tmp.reserve(count);            int idx1 = first;            int idx2 = mid;            int semicount = 0;            while ((idx1 < mid) || (idx2 < last)) {                if ((idx2 == last) || ((idx1 < mid) &&                       (nums[indices[idx1]] <= nums[indices[idx2]]))) {tmp.push_back(indices[idx1]);                    results[indices[idx1]] += semicount;                    ++idx1;                } else {tmp.push_back(indices[idx2]);                    ++semicount;                    ++idx2;                }            }            move(tmp.begin(), tmp.end(), indices.begin()+first);        }    }public:    vector<int> countSmaller(vector<int>& nums) {        int n = nums.size();        vector<int> results(n, 0);        vector<int> indices(n, 0);        iota(indices.begin(), indices.end(), 0);        merge_countSmaller(indices, 0, n, results, nums);        return results;    }};