4. Median of Two Sorted Arrays

二有序数组常用二分法, 要判断数组是否越界, 而数组的题常用递归, 擦半的方法, 比快排的方法, 少一个partition 遍历, 但是还是分成logn 层, Lintcode: Nuts & Bolts Problem

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

关键在于找到合适的比较位置, 准备擦出哪个数组的一半, 防止数组越界

The key point of this problem is to ignore half part of A and B each step recursively by comparing the median of remaining A and B:

if (aMid < bMid) Keep [aRight + bLeft]    
else Keep [bRight + aLeft]

As the following: time=O(log(m + n))

public double findMedianSortedArrays(int[] A, int[] B) {
	    int m = A.length, n = B.length;
	    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;
	}

public double getkth(int[] A, int aStart, int[] B, int bStart, int k) {
	if (aStart > A.length - 1) return B[bStart + k - 1];            
	if (bStart > B.length - 1) return A[aStart + k - 1];                
	if (k == 1) return Math.min(A[aStart], B[bStart]);
	
	int aMid = Integer.MAX_VALUE, bMid = Integer.MAX_VALUE;
	if (aStart + k/2 - 1 < A.length) aMid = A[aStart + k/2 - 1]; 
	if (bStart + k/2 - 1 < B.length) bMid = B[bStart + k/2 - 1];        
	
	if (aMid < bMid) 
	    return getkth(A, aStart + k/2, B, bStart,       k - k/2);// Check: aRight + bLeft 
	else 
	    return getkth(A, aStart,       B, bStart + k/2, k - k/2);// Check: bRight + aLeft
}

这是递归, 按照递归的写法来写, 递归出口, 越界出口和正常出口, 递归输入元素的续写(分情况一直擦最小的k/2), 

数组的递归, 常考虑其起始值和结束值. 比较值

 https://leetcode.com/problems/median-of-two-sorted-arrays/discuss/2471/Very-concise-O(log(min(MN)))-iterative-solution-with-detailed-explanation