36 数组中的逆序对

在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数P。并将P对1000000007取模的结果输出。 即输出P%1000000007

O(nlogn)

归并排序的思想

比如 5 7
4 6

5>4 那么比5大的数也会大于4
所以 cnt += m - i + 1

java:

public class Solution {
    private long cnt = 0 ;
    private int[] tmp ;
    public int InversePairs(int [] array) {
        tmp = new int[array.length] ;
        mergeSort(array , 0 , array.length - 1) ;
        return (int)(cnt % 1000000007) ;
    }
    
    private void mergeSort(int [] array , int left , int right){
        if (right - left < 1)
            return ;
        int mid = left + (right - left) / 2 ;
        mergeSort(array , left , mid) ;
        mergeSort(array , mid + 1 , right) ;
        merge(array , left , mid , right) ;
    }
    
    private void merge(int [] array , int left , int m , int right){
        int i = left , j = m+1 , k = left ;
        while(i <= m || j <= right){
            if (i > m){
                tmp[k++] = array[j++] ;
            }else if(j > right){
                tmp[k++] = array[i++] ;
            }else if(array[i] < array[j]){
                tmp[k++] = array[i++] ;
            }else{
                tmp[k++] = array[j++] ;
                cnt += m - i + 1 ;  // a[i] > a[j]，说明 a[i...mid] 都大于 a[j]
            }
        }
        for(k = left ; k <= right ; k++){
            array[k] = tmp[k] ;
        }
    }
}

C++:

vector<int>& data

class Solution {
private:
    long cnt = 0 ;
    vector<int> temp ;
public:
    int InversePairs(vector<int> data) {
        temp.resize(data.size()) ;
        mergeSort(data , 0 , data.size() - 1) ;
        return (int)(cnt%1000000007) ;
    }
    
    void mergeSort(vector<int>& data , int left , int right){
        if (right - left < 1){
            return ;
        }
        int mid = left + (right - left) / 2 ;
        mergeSort(data , left , mid) ;
        mergeSort(data , mid+1 , right) ;
        merge(data , left , mid , right) ;
    }
    
    void merge(vector<int>& data , int left ,int mid , int right){
        int i = left ;
        int j = mid+1 ;
        int k = left ;
        while(i <= mid || j <= right){
            if (i > mid){
                temp[k++] = data[j++] ;
            }else if (j > right){
                temp[k++] = data[i++] ;
            }else if (data[i] < data[j]){
                temp[k++] = data[i++] ;
            }else{
                temp[k++] = data[j++] ;
                cnt += mid-i+1 ;
            }
        }
        for(i = left ; i <= right ; i++){
            data[i] = temp[i] ;
        }
    }
};