leetcode:4. Median of Two Sorted Arrays（Java实现）

最新推荐文章于 2023-01-26 14:28:30 发布

JacobGo

最新推荐文章于 2023-01-26 14:28:30 发布

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本文介绍了一个LeetCode上的经典算法问题——如何找到两个已排序数组的中位数，且要求整体运行时间复杂度为O(log(m+n))。通过提供的Java代码示例，详细解释了使用分治策略来解决这个问题的方法。

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leetcode测试地址：https://leetcode.com/problems/median-of-two-sorted-arrays/#/description

There are two sorted arrays nums1 and nums2 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)).

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

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Java code：

package go.jacob.day617;

public class Demo2 {
	public double findMedianSortedArrays(int[] A, int[] B) {
		int m = A.length, n = B.length;
		// 不论总数是奇数还是偶数，以l和r为下标的两数的均值都是medium
		int l = (m + n + 1) / 2;
		int r = (m + n + 2) / 2;

		return (getkth(A, 0, B, 0, l) + getkth(A, 0, B, 0, r)) / 2.0;
	}

	private int getkth(int[] A, int aStart, int[] B, int bStart, int k) {
		if (aStart >= A.length)
			return B[bStart + k - 1];
		if (bStart >= B.length)
			return A[aStart + k - 1];
		if(k==1)
			return Math.min(A[aStart], B[bStart]);
		int aMin=Integer.MAX_VALUE,bMin=Integer.MAX_VALUE;
		if(aStart+k/2-1<A.length)
			aMin=A[aStart+k/2-1];
		if(bStart+k/2-1<B.length)
			bMin=B[bStart+k/2-1];
		
		if(aMin<bMin)
			return getkth(A,aStart+k/2,B,bStart,k-k/2);
		else
			return getkth(A,aStart,B,bStart+k/2,k-k/2);
	}
}