LeetCode | Median of Two Sorted Arrays-优快云博客

本文链接：https://blog.youkuaiyun.com/perfect8886/article/details/23748461

本文探讨了如何高效地找到两个已排序数组的中位数，使用了二分查找方法，确保了时间复杂度为 O(log(m+n))。通过转换为找第 K 大的数问题并小心处理边界条件，实现了解决方案。

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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)).

分析

转换为找第K大的数，需要十分小心边界条件。

代码

public class MedianOfTwoSortedArrays {
	public double findMedianSortedArrays(int A[], int B[]) {
		int M = A.length;
		int N = B.length;
		int total = M + N;
		if (total % 2 == 0) {
			return (solve(A, B, total / 2, 0, M - 1, 0, N - 1) + solve(A, B,
					total / 2 + 1, 0, M - 1, 0, N - 1)) / 2;
		} else {
			return solve(A, B, total / 2 + 1, 0, M - 1, 0, N - 1);
		}
	}

	private double solve(int A[], int B[], int k, int aStart, int aEnd,
			int bStart, int bEnd) {
		if (aEnd - aStart > bEnd - bStart) {
			return solve(B, A, k, bStart, bEnd, aStart, aEnd);
		}
		if (aEnd < aStart) {
			return B[k - 1];
		}
		if (k == 1) {
			return Math.min(A[aStart], B[bStart]);
		}
		int partASize = Math.min(k / 2, aEnd - aStart + 1);
		int pA = aStart + partASize;
		int pB = bStart + k - partASize;
		if (A[pA - 1] < B[pB - 1]) {
			return solve(A, B, k - partASize, pA, aEnd, bStart, bEnd);
		} else if (A[pA - 1] > B[pB - 1]) {
			return solve(A, B, partASize, aStart, aEnd, bStart + k - partASize,
					bEnd);
		} else {
			return A[pA - 1];
		}
	}
}