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  • Median of Two Sorted Arrays-----LeetCode

    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)).

    解题思路:

    该题可以解决所有求有序数组A和B有序合并之后第k小的数!

    该题的重要结论:

    如果A[k/2-1]<B[k/2-1],那么A[0]~A[k/2-1]一定在第k小的数的序列当中,可以用反证法证明。

    具体的分析过程可以参考http://blog.csdn.net/zxzxy1988/article/details/8587244

    class Solution {
    public:
        double findKth(int A[], int m, int B[], int n, int k)
        {
            //m is equal or smaller than n
            if (m > n)
                return findKth(B, n, A, m, k);
            if (m == 0)
                return B[k-1];
            if (k <= 1)
                return min(A[0], B[0]);
    
            int pa = min(k / 2, m), pb = k - pa;
            if (A[pa-1] < B[pb-1])
            {
                return findKth(A + pa, m - pa, B, n, k - pa);
            }
            else if(A[pa-1] > B[pb-1])
            {
                return findKth(A, m, B + pb, n - pb, k - pb);
            } else
                return A[pa-1];
        }
    
        double findMedianSortedArrays(int A[], int m, int B[], int n) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            int k = m + n;
            if (k & 0x1)
            {
                return findKth(A, m, B, n, k / 2 + 1);
            } else
            {
                return (findKth(A, m, B, n, k / 2) + findKth(A, m, B, n, k / 2 + 1)) / 2;
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/dollarzhaole/p/3153338.html
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