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:

 1 public class Solution {
 2     private long cnt = 0 ;
 3     private int[] tmp ;
 4     public int InversePairs(int [] array) {
 5         tmp = new int[array.length] ;
 6         mergeSort(array , 0 , array.length - 1) ;
 7         return (int)(cnt % 1000000007) ;
 8     }
 9     
10     private void mergeSort(int [] array , int left , int right){
11         if (right - left < 1)
12             return ;
13         int mid = left + (right - left) / 2 ;
14         mergeSort(array , left , mid) ;
15         mergeSort(array , mid + 1 , right) ;
16         merge(array , left , mid , right) ;
17     }
18     
19     private void merge(int [] array , int left , int m , int right){
20         int i = left , j = m+1 , k = left ;
21         while(i <= m || j <= right){
22             if (i > m){
23                 tmp[k++] = array[j++] ;
24             }else if(j > right){
25                 tmp[k++] = array[i++] ;
26             }else if(array[i] < array[j]){
27                 tmp[k++] = array[i++] ;
28             }else{
29                 tmp[k++] = array[j++] ;
30                 cnt += m - i + 1 ;  // a[i] > a[j]，说明 a[i...mid] 都大于 a[j]
31             }
32         }
33         for(k = left ; k <= right ; k++){
34             array[k] = tmp[k] ;
35         }
36     }
37 }

 

C++:

vector<int>& data

 1 class Solution {
 2 private:
 3     long cnt = 0 ;
 4     vector<int> temp ;
 5 public:
 6     int InversePairs(vector<int> data) {
 7         temp.resize(data.size()) ;
 8         mergeSort(data , 0 , data.size() - 1) ;
 9         return (int)(cnt%1000000007) ;
10     }
11     
12     void mergeSort(vector<int>& data , int left , int right){
13         if (right - left < 1){
14             return ;
15         }
16         int mid = left + (right - left) / 2 ;
17         mergeSort(data , left , mid) ;
18         mergeSort(data , mid+1 , right) ;
19         merge(data , left , mid , right) ;
20     }
21     
22     void merge(vector<int>& data , int left ,int mid , int right){
23         int i = left ;
24         int j = mid+1 ;
25         int k = left ;
26         while(i <= mid || j <= right){
27             if (i > mid){
28                 temp[k++] = data[j++] ;
29             }else if (j > right){
30                 temp[k++] = data[i++] ;
31             }else if (data[i] < data[j]){
32                 temp[k++] = data[i++] ;
33             }else{
34                 temp[k++] = data[j++] ;
35                 cnt += mid-i+1 ;
36             }
37         }
38         for(i = left ; i <= right ; i++){
39             data[i] = temp[i] ;
40         }
41     }
42 };

 

posted @ 2018-05-10 16:23  __Meng  阅读(133)  评论(0)  编辑  收藏  举报