4. Median of Two Sorted Arrays
官方的链接：4. 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)).
Example1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
问题描述
给定长度分别为m和n的两个排序数组nums1和nums2，在时间复杂度O(log (m+n))内算出数组的中位数
思路
两个有序数组的中位数和Top K问题类似。这里从小到大直接定位第k个，简单理解为从nums1中获取第i个，而从nums2中获取第j=k-i个，其中i=0~k。当定位到nums[i-1]<=nums2[j]和nums[j-1]<=nums[i]即完成，当然还有边界问题。
把中位数也当作是top k问题，最后进行奇偶判断。详情可参考这里Share my O(log(min(m,n)) solution with explanation，代码也是借鉴的。
值得注意的是几个边界判断问题，比如i为0或者m，j为0或者n。
1 public class Q4_MedianOfTwoSortedArrays { 2 public double findMedianSortedArrays(int[] nums1, int[] nums2) { 3 4 int m = nums1.length; 5 int n = nums2.length; 6 // make sure that m <= n 7 if (m > n) { 8 return findMedianSortedArrays(nums2, nums1); 9 } 10 // n>=m,i = 0 ~ m,so j = (m + n + 1) / 2 -i > 0. 11 int i = 0, j = 0, imax = m, imin = 0, halfLen = (m + n + 1) / 2; 12 int maxLeft = 0, minRight = 0; 13 while (imin <= imax) { 14 i = (imin + imax) / 2; 15 j = halfLen - i; 16 if (i < m && nums2[j - 1] > nums1[i]) { 17 imin = i + 1; 18 } else if (i > 0 && nums1[i - 1] > nums2[j]) { 19 imax = i - 1; 20 } else { 21 if (i == 0) { 22 //the target is in nums2 23 maxLeft = nums2[j - 1]; 24 } else if (j == 0) { 25 //the target is in nums1 26 maxLeft = nums1[i - 1]; 27 } else { 28 maxLeft = Math.max(nums1[i - 1], nums2[j - 1]); 29 } 30 break; 31 } 32 } 33 //odd 34 if ((m + n) % 2 == 1) { 35 return (double)maxLeft; 36 } 37 //even 38 if (i == m) { 39 //nums1 is out of index m 40 minRight = nums2[j]; 41 } else if (j == n) { 42 //nums2 is out of index n 43 minRight = nums1[i]; 44 } else { 45 //others 46 minRight = Math.min(nums1[i], nums2[j]); 47 } 48 return (double) (maxLeft + minRight) / 2; 49 } 50 51 public static void main(String[] args) { 52 new Q4_MedianOfTwoSortedArrays().findMedianSortedArrays(new int[] { 1, 3 }, new int[] { 2 }); 53 54 } 55 }
