[LeetCode] Median of Two Sorted Arrays [16]

题目

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)).

原题链接（点我）

解题思路

返回两个排序数组的中位数。这个题可以有以下几个思路：
首先可以想到的是将两个数组merge起来，然后返回其中位数。
第二个是，类似merge的思想加上计数，找到(m+n)/2个数或者其前后的数，这个就可以算出中位数。这个方法对于各种情况需要一一考虑到。
第三个，假设A[k/2-1]<B[k/2-1]，那么A[k/2-1]之前的数一定在整个有序数列中(m+n)/2之前。
这里我给出后面两种思路的代码。

代码实现

代码一（思路三）

class Solution {
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        int len = m+n;
        if(len & 0x1){
            return findkth(A, m, B, n, len/2 +1);
        }else{
            return ( findkth(A, m, B, n, len/2) + findkth(A, m, B, n, len/2+1) )/2.0;
        }
    }
    
    double findkth(int A[], int m, int B[], int n, int k){
        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 Aindex = min(m, k/2);
        int Bindex = k-Aindex;
        if(A[Aindex-1] < B[Bindex-1]){
            return findkth(A+Aindex, m-Aindex, B, n, k-Aindex);
        }
        else if(A[Aindex-1] > B[Bindex-1]){
            return findkth(A, m, B+Bindex, n-Bindex, k-Bindex);
        }
        else{
            return A[Aindex-1];
        }
    }
};

代码二（思路二--计数）

class Solution {
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        int ret = -1;
        if((A==NULL && B==NULL) || m<0 || n<0)
            return (double)ret;
        if(A==NULL || m<=0){
            if(n & 1)
                return (double)B[n/2];
            else{
                return (double)(B[n/2]+B[n/2-1]) / 2.0;
            }
        }
        if(B==NULL || n<=0){
            if(m & 1)
                return (double)A[m/2];
            else{
                return (double)(A[m/2]+A[m/2-1]) / 2.0;
            }
        }
        int mid = (m+n)/2;
        int pre = 0;
        int i=0, j=0;
        int count=0;
        int token = 0;
        while(i<m && j<n){
            if(A[i]<=B[j]){
                if(count!=mid)
                    pre = A[i];
                if(count==mid){
                    token = A[i];++count;
                    break;
                }
                ++i;
            }else{
                if(count!=mid)
                    pre = B[j];
                if(count==mid){
                    token = B[j];++count;
                    break;
                }
                ++j;
            }
            ++count;
        }
        if(count<=mid){
            if(i<m){
                if(count==mid){
                    token = A[i];
                }else{
                    while(count<mid){
                        pre = A[i++];
                        ++count;
                    }
                    token = A[i];
                }
            }
            if(j<n){
                if(count==mid){
                    token = B[j];
                }else{
                    while(count<mid){
                        pre = B[j++];
                        ++count;
                    }
                    token = B[j];
                }
            }
        }
        
        if((m+n) & 1 ){
            //odd
            return (double)(token);
        }
        else{
            //even
            return (double)(pre+token)/2.0;
        }
    }
};

参考

http://blog.csdn.net/yutianzuijin/article/details/11499917

-----------------------------------------------------------------------------------------------------------------------------------

[LeetCode] Median of Two Sorted Arrays [16],布布扣,bubuko.com

[LeetCode] Median of Two Sorted Arrays [16]

原文：http://blog.csdn.net/swagle/article/details/29197389