数组中的逆序对


http://www.acmerblog.com/offer09-2527.html
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

直接的做法是逐个统计,复杂度是N^2,可以利用归并排序的思想,在排序过程中统计逆序对的个数。
时间复杂度依然是 N*Log(N)。 可以从代码中看到,只是比归并排序多了一句代码:cnt += (end-j+1).
由于最终个数可能超过int,这里用long long
13int n, arr[100010], tmp[100010];
14 
15//归并排序,过程中 统计逆序数
16ll merge(int start, int mid, int end){
17    ll cnt = 0;
18    int i = start, j = mid+1, k = start;
19    while( i<=mid && j<= end){
20        //从大到小排序
21        if(arr[i] > arr[j]){
22            cnt += (end-j+1); //右面剩下的都是逆序
23            tmp[k++] = arr[i++];
24        }else{
25            tmp[k++] = arr[j++];
26        }
27    }
28    while(i<=mid) tmp[k++] = arr[i++];
29    while(j<=end) tmp[k++] = arr[j++];
30    for(int i=start; i<=end; i++) arr[i] = tmp[i];
31    return cnt;
32}
33ll inversePairs(int start, int end){
34    ll cnt = 0;
35    if(start < end){
36        int mid = (start + end)/2;
37        cnt += inversePairs(start, mid); //左半部分 逆序对数量
38        cnt += inversePairs(mid+1, end); //右半部分
39        cnt += merge(start, mid, end); //合并两部分,并计算数量
40    }
41    return cnt;
42}
43int main() {
44    //freopen("in.txt", "r", stdin);
45    whilescanf("%d", &n) != EOF){
46        for(int i=0; i<n; i++) scanf("%d", &arr[i]);
47        ll ans = inversePairs(0, n-1);
48        printf("%lld\n",ans);
49    }
50    return 0;
51}

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