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  • 求逆序对

    class Solution {
    public:
        int result = 0;
    
        int InversePairs(vector<int> data) {
    
            int len = data.size();
            vector<int> temp(len);
            MergeSort(data, temp, 0, len-1);
            return result;
        }
    
        void MergeSort(vector<int>& data, vector<int>& temp, int begin, int end)
        {
            if (begin < end)
            {
                int mid = (end - begin) / 2 + begin;
                MergeSort(data, temp, begin, mid);
                MergeSort(data, temp, mid+1, end);
                MergeArray(data, temp, begin, mid, end);
            } 
        }
    
        void MergeArray(vector<int>& data, vector<int>& temp, int begin, int mid, int end)
        {
            int i = begin;
            int j = mid + 1;
            int k = 0;
    
            while (i <= mid && j <= end)
            {
                if (data[i] < data[j])
                {
                    temp[k++] = data[i++];
                }
                // 若左半部分当前元素大于右半部分当前元素
                // 则左半部分当前元素后面的每个值都大于它
                else
                {
                    result += (mid - i + 1);
                    result %=  1000000007;
                    temp[k++] = data[j++];
                }
            }
    
            while (i <= mid)
            {
                temp[k++] = data[i++];
            }
    
            while (j <= end)
            {
                temp[k++] = data[j++];
            }
    
            for (i = 0; i < k; i++)
            {
                data[begin+i] = temp[i];
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/dealer/p/13805682.html
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