2017-09-15 LeetCode_493 Reverse Pairs

493. Reverse Pairs

Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].

You need to return the number of important reverse pairs in the given array.

Example1:

Input: [1,3,2,3,1]Output: 2

Example2:

Input: [2,4,3,5,1]Output: 3

Note:

1. The length of the given array will not exceed 50,000.
2. All the numbers in the input array are in the range of 32-bit integer.

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 ((long long)nums[i] > 2*(long long)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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    int reversePairs(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 0;

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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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        int myans = 0; 

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

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            final[d[i]] = ans[i]; myans += ans[i];

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        }

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        return myans;

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    }

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};