LeeCode(4) Median of Two Sorted ArraysC语言版本

题目： 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

思路： 由于数组是已经排序好的数组，则直接合并数组，然后根据两数组总的元素个数之和，获得中间数字。其中可以做一些优化。

C语言代码：

double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size){ double result =0; int sumSize = nums1Size + nums2Size; int len = sumSize / 2; int size =len+1; int * merge =(int *)malloc (size * sizeof(int)); int i,n1,n2; for(i=0,n1 =0 ,n2 =0 ; i < size ; i++) { if(n1<nums1Size && n2<nums2Size) { if(nums1[n1] > nums2[n2]) { merge[i] = nums2[n2]; n2++; }else { merge[i] =nums1[n1]; n1++; } } else if(n1 >= nums1Size) { merge[i] = nums2[n2]; n2++; } else if(n2 >= nums2Size) { merge[i] =nums1[n1]; n1++; } } if(sumSize %2 !=0) { result= merge[len]; } else { result = merge[len-1] + merge[len]; result = result /2; } return result; }

效率： Leecode中同一代码执行的时间可能不同。本系列中，取代码最佳效率。