Counting Inversion Pairs in an Array

最新推荐文章于 2025-01-16 16:47:57 发布

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本文介绍了一种使用Merge Sort算法在O(nlog(n))时间复杂度内计算数组中逆序对数量的方法。通过将数组分为两部分并递归地计算每个部分的逆序对，然后合并时统计跨越两个部分的逆序对，最终得到整个数组的逆序对总数。

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Given an array, for example, 2 4 6 1 3 5, an inversion pair is the pair whose first value is larger than its second value according to the sequence from left to right, (2,1) (4,1) (4,3) (6,1) (6,3) (6,5). The following code is an O(nlog(n)) algorithm based on merge sort to count the inversion pairs in an array.

long long inversionCount(vector<int> &A, int l, int r) {
	if (l >= r) return 0;
	int m = (l + r) >> 1;
	long long res = inversionCount(A, l, m);
	res += inversionCount(A, m+1, r);
	vector<int> B(A.begin()+l, A.begin()+r+1);
	int i = l, j = m + 1, k = l;
	while (i<=m && j<=r) {
		if (B[i-l] <= B[j-l]) {
			A[k++] = B[(i++)-l];
		} else {
			A[k++] = B[(j++)-l];
			res += m - i + 1;
		}
	}
	while (i <= m) A[k++] = B[(i++)-l];
	while (j <= r) A[k++] = B[(j++)-l];
	return res;
}