问题描述：
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)).
解题思路：
看到时间复杂度的时候就知道这种应该使用二分查找法了，否则如果实现log的时间复杂度？
思路已经有大神提供了，说的非常清楚，附上链接地址：http://my.oschina.net/jdflyfly/blog/283267
代码如下：
1 public class Solution { 2 public double findMedianSortedArrays(int[] nums1, int[] nums2) { 3 int total = nums1.length + nums2.length; 4 if (total % 2 == 0) 5 return (findKth(nums1, 0, nums1.length - 1, nums2, 0, 6 nums2.length - 1, total / 2) + findKth(nums1, 0, 7 nums1.length - 1, nums2, 0, nums2.length - 1, total / 2 + 1)) / 2; 8 else 9 return findKth(nums1, 0, nums1.length - 1, nums2, 0, 10 nums2.length - 1, total / 2 + 1); 11 12 } 13 14 public double findKth(int[] a, int astart, int aend, int[] b, 15 int bstart, int bend, int k) { 16 if (aend - astart > bend - bstart) 17 return findKth(b, bstart, bend, a, astart, aend, k); 18 if (astart > aend) 19 return b[k - 1]; 20 if (k == 1) 21 return a[astart] > b[bstart] ? b[bstart] : a[astart]; 22 else { 23 int la = Math.min(k / 2, aend - astart + 1); 24 int lb = k - la; 25 if (a[astart + la - 1] == b[bstart + lb - 1]) 26 return a[astart + la - 1]; 27 else if (a[astart + la - 1] < b[bstart + lb - 1]) 28 return findKth(a, astart + la, aend, b, bstart, bend, k - la); 29 else 30 return findKth(a, astart, aend, b, bstart + lb, bend, k - lb); 31 } 32 33 } 34 }
Java [leetcode 4] Median of Two Sorted Arrays
原文：http://www.cnblogs.com/zihaowang/p/4455798.html