标签：leetcode   算法   

问题描述：

There are two sorted arrays A and B 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)).

基本思想：

此题思想很简单，设定两个指针，从两个数组的开始一个个比较并适当的移动指针，直到找到中位数。

需要注意的一点是：中位数的定义。数据总和为偶数时，中位数如何处理？(在面试中，这个最好要对面试官问清楚。)此题的处理是采用两个中位数的平均值

代码：

 double findMedianSortedArrays(int A[], int m, int B[], int n) {  //C++
        int pre = 0;
        int midpos = (m+n)/2;
        bool isodd = (m+n)%2 == 1;
        
        if(n == 0)
        {
            if(isodd)
                return A[(m-1)/2];
            else 
                return ((double)A[(m-1)/2]+(double)A[(m-1)/2+1])/2;
            
        }
        if(m == 0)
        {
            if(isodd)
                return B[(n-1)/2];
            else 
                return ((double)B[(n-1)/2]+(double)B[(n-1)/2+1])/2;
        }    
        
        int pa = 0, pb = 0;
        double median;
        int nextmed = 0;
        if(isodd)
        {
            while(pre != midpos+1 && pa < m&& pb < n )
            {
                
                if(A[pa] < B[pb])
                {
                    median = A[pa];
                    pa++;
                }
                else
                {
                    median = B[pb];
                    pb++;
                }
                pre++;
            }
            if(pre < midpos+1)
            {
            
                if(pa == m)
                    return B[midpos-pre+pb];
                else return A[midpos-pre+pa];
            }
            return median;
        }
        else
        {
            while(pre != midpos && pa < m&& pb < n )
            {
                
                if(A[pa] < B[pb])
                {
                    median = A[pa];
                    pa++;
                }
                else
                {
                    median = B[pb];
                    pb++;
                }
                pre++;
            }
            if(pre < midpos)
            {
            
                if(pa == m)
                    return ((double)B[midpos-pre+pb-1]+(double)B[midpos-pre+pb])/2;
                else return ((double)A[midpos-pre+pa-1]+(double)A[midpos-pre+pa])/2;
            }
            else
            {
                if(pa == m)
                    return ((double)median + B[pb])/2;
                else if(pb == n)
                    return ((double)median + A[pa])/2;
                else {
                    nextmed = (A[pa] > B[pb])? B[pb]:A[pa];
                    return ((double)median + nextmed)/2;
                }
            }
        }
    }

[leetcode]Median of Two Sorted Arrays

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/41410389