LeetCode315. Count of Smaller Numbers After Self

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315. Count of Smaller Numbers After Self

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.

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题目：求解nums[i]右侧元素中比nums[i]小的元素的个数count[i] 。

思路：参见solution using merge sort with explanation。题目的本质上同LeetCode493. Reverse Pairs。利用归并排序，求nums[i]右侧元素中比nums[i]本身小的元素的个数，相当于在并的过程中，左侧数组nums[i]合并到数组merge之前，发生了由右侧数组元素先添加到merge中的次数。这里有个技巧，我们排序的过程中，不改变元素数组nums，而是改变nums[i]对应的下标。

工程代码下载

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n), index(n);
        for(int i=0; i < n; ++i){
            index[i] = i;
        }
        countSmaller(nums, index, res, 0, n-1);
        return res;
    }
private:
    void countSmaller(vector<int>& nums, vector<int>& index, vector<int>& res, int l, int r){
        if(l >= r) return;

        int m = l + (r - l)/2;
        countSmaller(nums, index, res, l, m);
        countSmaller(nums, index, res, m+1, r);

        vector<int> merge(r-l+1);
        int i = l, j = m + 1, k = 0, p = 0;
        while(i <= m && j <=r){
            if(nums[index[i]] > nums[index[j]]){
                merge[k++] = index[j++];
                p++;
            }
            else{
                res[index[i]] += p;
                merge[k++] = index[i++];
            }
        }

        while(i <= m){
            res[index[i]] += p;
            merge[k++] = index[i++];
        }

        while(j <= r){
            merge[k++] = index[j++];
        }

        copy(merge.begin(), merge.end(), index.begin() + l);
    }
};