标签：EDA res ted example 讲解 and lse val div
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)).
You may assume nums1 and nums2 cannot be both empty.
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
class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { int m = nums1.length, n = nums2.length, left = (m + n + 1) / 2, right = (m + n + 2) / 2; return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0; } int findKth(int[] nums1, int i, int[] nums2, int j, int k) { if (i >= nums1.length) return nums2[j + k - 1]; if (j >= nums2.length) return nums1[i + k - 1]; if (k == 1) return Math.min(nums1[i], nums2[j]); int midVal1 = (i + k / 2 - 1 < nums1.length) ? nums1[i + k / 2 - 1] : Integer.MAX_VALUE; int midVal2 = (j + k / 2 - 1 < nums2.length) ? nums2[j + k / 2 - 1] : Integer.MAX_VALUE; if (midVal1 < midVal2) { return findKth(nums1, i + k / 2, nums2, j, k - k / 2); } else { return findKth(nums1, i, nums2, j + k / 2, k - k / 2); } } }
hard名不虚传，是真的hard。不过思想可以看懂，代码实现还是有点困难才能理解，记下来先。
https://leetcode.windliang.cc/leetCode-4-Median-of-Two-Sorted-Arrays.html
http://www.cnblogs.com/grandyang/p/4465932.html
两位大佬的讲解，之后细细品尝
4. Median of Two Sorted Arrays
标签：EDA res ted example 讲解 and lse val div
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/10353632.html
