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)).

题意

有两个排序好的数组nums1 & nums2.找到这两个数组的中位数。

思路

令sum等于两个数组大小的和。

如果sum==1,返回这个数。

否则，如果sum是偶数，返回这两个数组中第sum/2大的数和第sum/2+1大的数的均值；

如果sum是奇数，返回这两个数组中第sum/2+1大的数。

 1 double find(int* nums1, int nums1Size, int* nums2, int nums2Size,int t){ //返回两个数组中第t小的数
 2     int pos1=0,pos2=0;
 3     int ans;
 4     while(t--)
 5     {
 6         if(pos1<nums1Size && pos2<nums2Size)
 7         {
 8             if(nums1[pos1]>nums2[pos2])
 9                ans=nums2[pos2++];
10             else
11                ans=nums1[pos1++];
12         }
13         else if(pos1<nums1Size)
14             ans=nums1[pos1++];
15         else if(pos2<nums2Size)
16             ans=nums2[pos2++];
17     }
18     return ans;
19 }
20 double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
21     int sum=nums1Size+nums2Size;
22     if(sum == 1)
23     {
24         if(nums1Size == 1)
25             return nums1[0];
26         else return nums2[0];
27     }
28     else
29     {
30         if(sum%2==0)
31             return (find(nums1,nums1Size,nums2,nums2Size,sum/2)+find(nums1,nums1Size,nums2,nums2Size,sum/2+1))/2;
32          else
33             return find(nums1,nums1Size,nums2,nums2Size,sum/2+1);
34     }
35 }

转载于:https://www.cnblogs.com/madking/p/4657690.html

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题意

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