[LeetCode]4.Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and nrespectively.Find the median of the two sorted arrays. The overall run timecomplexity 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难点主要在于复杂度的要求，logn的复杂度我们就会想到二分查找，二分查找本质上就是对一个有序值的寻找过程，会比遍历的寻找合适值寻找复杂度更低，两个数组都做二分查找的值并没什么意义，我们需要想到一个隐藏条件oooxxxooooxxxx寻找中位数其实就是剔除比较小的数，而这些小的数都在整个数组的左边，且个数是固定的。public double findMedianSortedArrays(int[] nums1, int[] nums2) { int l1=nums1.length; int l2=nums2.length; if(l1>l2) return findMedianSortedArrays(nums2,nums1); int left=0; int right=l1; int need=(l1+l2-1)/2; while(left<=right){ int count1=(left+right)/2; int count2=need-count1; if(count1<l1 && count2>=1 && nums1[count1]<nums2[count2-1]) left=count1+1; else if(count1>=1 && nums2[count2]<nums1[count1-1]) right=count1-1; else{ if((l1+l2)%2==1){ int[] array=new int[]{count1>=l1?Integer.MAX_VALUE:nums1[count1],count2>=l2?Integer.MAX_VALUE:nums2[count2]}; Arrays.sort(array); return (double) (array[0]); }else{ int[] array=new int[]{count1>=l1?Integer.MAX_VALUE:nums1[count1],count2>=l2?Integer.MAX_VALUE:nums2[count2],count1+1>=l1?Integer.MAX_VALUE:nums1[count1+1],count2+1>=l2?Integer.MAX_VALUE:nums2[count2+1]}; Arrays.sort(array); return (double) (array[0]+array[1])/2.0; } } } return 0.0;}