4. Median of Two Sorted Arrays

最新推荐文章于 2023-06-12 16:30:30 发布

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最新推荐文章于 2023-06-12 16:30:30 发布

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本文链接：https://blog.youkuaiyun.com/yang10560/article/details/80558828

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本文介绍了一种在O(log(m+n))的时间复杂度内找到两个有序数组中位数的方法。通过合并两个数组并选取中间元素实现，适用于算法面试和技术实践。

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转 https://blog.youkuaiyun.com/yuxin6866/article/details/52425202

**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)).**

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

double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    int mid = (nums1Size+nums2Size)/2;
    int j,k;
    j=0;k=0;//cursor
    int *p = (int *)malloc(sizeof(int)*(mid+1));

    for(int i = 0 ; i < mid+1; i++){
        if((k>=nums2Size||nums1[j]<nums2[k]) && j<nums1Size){
            *(p+i) = nums1[j++];
        }else{
             *(p+i) = nums2[k++];
        }
    }


    if ((nums1Size+nums2Size)%2 != 0){  
            return (double)(*(p+mid));  
        }else{  
            return (double)(*(p+mid-1)+*(p+mid))/2;  
    }  
}