题解 | 数组中的逆序对

 
class Solution {
    const int MOD = 1000000007;
    int count = 0;

    void mergeSort(vector<int>& nums, int left, int right) {
        if (left >= right) return;
        
        int mid = left + (right - left) / 2;
        mergeSort(nums, left, mid);
        mergeSort(nums, mid + 1, right);
        merge(nums, left, mid, right);//对交叉归并的理解更加重要
    }

    void merge(vector<int>& nums, int left, int mid, int right) {
        vector<int> temp(right - left + 1);
        //先定义一个临时数组
        int i = left, j = mid + 1, k = 0;
        
        while (i <= mid && j <= right) {
            if (nums[i] <= nums[j]) {
                temp[k++] = nums[i++];
            } else {
                // 当nums[i] > nums[j]时，nums[i...mid]都会比nums[j]大
                count = (count + mid - i + 1) % MOD;
                temp[k++] = nums[j++];
            }
        }
        //遍历，不进行排序
        
        while (i <= mid) temp[k++] = nums[i++];
        while (j <= right) temp[k++] = nums[j++];
        
        for (int p = 0; p < temp.size(); p++) {
            nums[left + p] = temp[p];
        }
    }

public:
    int InversePairs(vector<int>& nums) {
        if (nums.empty()) return 0;
        mergeSort(nums, 0, nums.size() - 1);
        return count;
    }
};

暂时没看懂交叉归并这部分