2017-09-14 LeetCode_315 Count of Smaller Numbers After Self

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:

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].

solution:

class Solution {

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public:

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    void divide(vector<int>& nums, vector<int>& ans, vector<int>& d, vector<vector<int> >& temp, int start, int end) {

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        if (start == end) return;

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        divide(nums, ans, d, temp, start, (start+end)/2);

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        divide(nums, ans, d, temp, (start+end)/2+1, end);

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        int i = start, j = (start+end)/2+1, k = start;

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        while (i <= (start+end)/2 || j <= end) {

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            if (i == (start+end)/2+1) j++;

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            else if (j == end+1) i++;

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            else if (nums[i] > nums[j]) {

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                ans[i] += end-j+1;

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                i++;

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            } else j++;

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        }

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        i = start, j = (start+end)/2+1;

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        while (i <= (start+end)/2 || j <= end) {

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            if (i == (start+end)/2+1) {

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                temp[0][k] = nums[j]; temp[1][k] = ans[j]; temp[2][k] = d[j];

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                k++; j++;

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            } else if (j == end+1) {

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                temp[0][k] = nums[i]; temp[1][k] = ans[i]; temp[2][k] = d[i];

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                k++; i++;

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            } else if (nums[i] > nums[j]) {

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                temp[0][k] = nums[i]; temp[1][k] = ans[i]; temp[2][k] = d[i];

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                k++; i++;

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            } else {

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                temp[0][k] = nums[j]; temp[1][k] = ans[j]; temp[2][k] = d[j];

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                k++; j++;

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            }

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        }

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        for (int l = start; l <= end; l++) {

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            nums[l] = temp[0][l]; ans[l] = temp[1][l]; d[l] = temp[2][l];

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        }

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    }

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    vector<int> countSmaller(vector<int>& nums) {

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        vector<int> ans, d, final, no;

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        vector<vector<int> > temp;

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        if (nums.size() == 0) return ans;

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        temp.push_back(no); temp.push_back(no); temp.push_back(no);

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        for (int i = 0; i < nums.size(); i++) {

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            ans.push_back(0); d.push_back(i); final.push_back(0);

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            temp[0].push_back(0); temp[1].push_back(0); temp[2].push_back(0);

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        }

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        divide(nums, ans, d, temp, 0, nums.size()-1);

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        for (int i = 0; i < nums.size(); i++) final[d[i]] = ans[i];

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        return final;

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    }

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};