题解 | 数组中的逆序对
数组中的逆序对
https://www.nowcoder.com/practice/96bd6684e04a44eb80e6a68efc0ec6c5
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
* @param nums int整型一维数组
* @return int整型
*/
private int merge_sort(int[] nums, int l, int r, int[] s) {
if (l >= r) return 0;
int mid = l + r >> 1;
int ans = 0;
ans += merge_sort(nums, l, mid, s) + merge_sort(nums, mid + 1, r, s);
int i = l, j = mid + 1, k = 0;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j]) {
s[k++] = nums[i++];
} else {
ans += mid - i + 1;
ans %= 1000000007;
s[k++] = nums[j++];
}
}
while (i <= mid) s[k++] = nums[i++];
while (j <= r) s[k++] = nums[j++];
for (int m = l, n = 0; m <= r; m++, n++) nums[m] = s[n];
return ans;
}
public int InversePairs(int[] nums) {
if (nums == null || nums.length == 0) return 0;
int[] s = new int[nums.length];
return merge_sort(nums, 0, nums.length - 1, s);
}
}


