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)).
两个有序数组，找中位数。
public class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { int len1 = nums1.length; int len2 = nums2.length; if(len1==0){ if(len2%2==1){ return (double)nums2[len2/2]; } else return ((double)nums2[(len2-1)/2]+nums2[len2/2])/2; } if(len2==0){ if(len1%2==1){ return (double)nums1[len1/2]; } else return ((double)nums1[(len1-1)/2]+nums1[len1/2])/2; } int mid = (len1+len2-1)/2; int midval1=0; int midval2=0; int i=0,j=0,k=0; for(;i<mid;i++){ if(j==len1){ k++; } else if(k==len2||nums1[j]<=nums2[k]){ j++; } else{ k++; } } if(j==len1){ midval1 = nums2[k]; midval2 = nums2[k+1]; } else if(k==len2){ midval1 = nums1[j]; midval2 = nums1[j+1]; } else{ midval1 = Math.min(nums1[j],nums2[k]); int m=(nums1[j]<=nums2[k])?j++:k++; if(j==len1) midval2=nums2[k]; else if(k==len2) midval2=nums1[j]; else midval2 = Math.min(nums1[j],nums2[k]); } if((len1+len2)%2==1) { return (double)midval1; } else return ((double)midval1+midval2)/2; } }这个代码写的有点冗长，勉强看吧。。 思路就是对两个数组逐个比较计数到mid，如果两个数组长度和是偶数就还要看mid+1