[leetcode]t002-Median of Two Sorted Arrays_leetcode t002-优快云博客

本文链接：https://blog.youkuaiyun.com/YangJSJ/article/details/42526275

本文详细介绍了如何使用二分查找算法解决两个已排序数组合并后的中位数查找问题，提供了递归和非递归两种实现方式，并分析了两者的时间复杂度。

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题目：There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

思路：题目直接要求O(logn)的时间复杂的，基本上可以直接考虑二分查找的思想．关键在于细节的实现．

递归版

public class Solution {
double findKth(int a[], int af, int alen, int b[], int bf, int blen, int k) {
		if (alen > blen)
			return findKth(b, bf, blen, a, af, alen, k);
		if (alen == 0)
			return b[k - 1];
		if (k == 1)
			return Math.min(a[af], b[bf]);
		int pa = Math.min(k / 2, alen), pb = k - pa;
		if (a[af + pa - 1] < b[bf + pb - 1])
			return findKth(a, af + pa, alen - pa, b, bf, blen, k - pa);
		else if (a[af + pa - 1] > b[bf + pb - 1])
			return findKth(a, af, alen, b, bf + pb, blen - pb, k - pb);
		else
			return a[af + pa - 1];
	}

	double findMedianSortedArrays(int A[], int B[]) {
		int m = A.length;
		int n = B.length;
		if ((m + n) % 2 == 1)
			return findKth(A, 0, m, B, 0, n, (m + n) / 2 + 1);
		else
			return (findKth(A, 0, m, B, 0, n, (m + n) / 2) + findKth(A, 0, m,
					B, 0, n, (m + n) / 2 + 1)) / 2;
	}

}

Runtime: 419 ms

非递归版

public class Solution {
	double findMedianSortedArrays(int A[], int B[]) {
		int m = A.length;
		int n = B.length;
		if ((m + n) % 2 == 1)
			return findKth(A, B, (m + n) / 2 + 1);
		else
			return (findKth(A, B, (m + n) / 2) + findKth(A, B, (m + n) / 2 + 1)) / 2;
	}

	double findKth(int a[], int b[], int k) {
		int af = 0, alen = a.length;
		int bf = 0, blen = b.length;
		int pa = 0, pb = 0;
		while (k > 0 ) {
			if (alen == 0)
				return b[k - 1];
			if (blen == 0)
				return a[k - 1];
			if (k == 1)
				return Math.min(a[af], b[bf]);
			if (alen > blen) {
				pb = Math.min(k / 2, blen);
				pa = k - pb;
				if (a[af + pa - 1] == b[bf + pb - 1])
					return b[bf + pb - 1];
			} else {
				pa = Math.min(k / 2, alen);
				pb = k - pa;
				if (a[af + pa - 1] == b[bf + pb - 1])
					return a[af + pa - 1];
			}
			if (a[af + pa - 1] < b[bf + pb - 1]) {
				af = af + pa;
				alen = alen - pa;
				k = k - pa;
			} else {
				bf = bf + pb;
				blen = blen - pb;
				k = k - pb;
			}
		}
		return -1;
	}
}

Runtime: 422 ms

时间上差不多，甚至久了些～～