leetCode: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.

解題思路：

1.歸併排序

用歸併排序將數組從小到大排序，在歸併過程中，將數組從前往後，從小到大合併，並用一個索引p記錄當前處於前一半歸併數組的nums[i]值有多少個後一半數組的nums[j]，使得nums[i]>2*nums[j]，然後在此基礎上往後面疊加p，因爲對於處於前一半數組的nums[i] <= nums[i']，必有p<=p'。故p的值只需要逐步往後累加。再將各個p值加起來就是所求答案。

class Solution {
public:
    int ans;
    void merge(vector<int>& nums, int L, int R){
        int len = R - L + 1, mid = (L + R)/2;
        int* c = new int[len];
        int i = L, j = mid + 1, k = 0;
        int p = j;
        while(i <= mid){
            while(p <= R && nums[i] > 2LL * nums[p])p++;
            ans += p - mid - 1;
            while(j <= R && nums[i] >= nums[j])c[k++] = nums[j++];
            c[k++] = nums[i++];
        }
        while(j <= R)c[k++] = nums[j++];
        for(i = 0; i < len; i++)nums[i + L] = c[i];
        delete[] c;
    }
    void mergeSort(vector<int>& nums, int L, int R){
        if(L >= R)return;
        int mid = (L + R)/2;
        mergeSort(nums, L, mid);
        mergeSort(nums, mid + 1, R);
        merge(nums, L, R);
    }
    int reversePairs(vector<int>& nums) {
        int len = nums.size();
        ans = 0;
        mergeSort(nums, 0, len - 1);
        /*
        for(int i = 0; i < len; i++)cout << nums[i] << " ";
        cout << endl;
        */
        return ans;
    }
};