题意:给一个数组,求出数组中存在的逆序对的个数,逆序对即两数中前面的数字大于后面的数字
思路:暴力枚举很简单,复杂度为O(n²),可以用归并排序的性质,归并后两个子数组中设前面为i,后面为j,若array[i]>array[j]
,则i到mid的数即为当前可知逆序对数(排序完后前面数组i到mid的数都比i位置的大,故排序前乱序时前面数组相对后面数组逆序对也有这么多),将所有子数组的逆序对统计即为数组的总逆序对个数
public class Solution {
public int InversePairs(int [] array) {
return solve(array, 0, array.length-1);
}
public int solve(int[] array, int l, int r) {
if (l >= r)
return 0;
int mid = (l+r)>>1;
int left = solve(array, l, mid);
int right = solve(array, mid+1, r);
return (left+right+merge(array, l, r))%1000000007;
}
public int merge(int[] array, int l, int r) {
int[] temp = new int[r-l+1];
int i = l, mid = (l+r)>>1, j = mid + 1, cnt = 0, cur = 0;
while (i <= mid || j <= r) {
if (j > r)
temp[cur++] = array[i++];
else if (i > mid)
temp[cur++] = array[j++];
else if (array[i] > array[j]) {
cnt = (cnt+mid-i+1)%1000000007;
temp[cur++] = array[j++];
} else {
temp[cur++] = array[i++];
}
}
for (int k = 0; k < temp.length; k++)
array[k+l] = temp[k];
return cnt;
}
}