c++-median-of-two-sorted-arrays

题目描述

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)).

也就是将两个数组合并为一个数组，然后找出这个新数组的中间值，如果是奇数个元素，中间值为中间，如果是偶数，则中间值为中间两个数的均值：

提供两种解法：

解法1：将两个数组由小到大合并为一个，然后找出中间值即可：

class Solution {
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        vector<int> res;
        int i=0;
        int j=0;
         
        if(m==0){
            if(n%2==0)
                return (double)(B[n/2]+B[n/2-1])/2;
            else
                return (double)(B[n/2]);
        }
        if(n==0){
            if(m%2==0)
                return (double)(A[m/2]+A[m/2-1])/2;
            else
                return (double)(A[m/2]);
        }
        while(i<m){
            while(j<n){
                if(A[i]>=B[j]){
                    res.push_back(B[j]);
                    j++;
                }
                else{
                    res.push_back(A[i]);
                    i++;
                    break;
                }
            }
            if(j==n && i<m){
                res.push_back(A[i]);
                i++;
            }
        }
        if(i==m && j<n){
            while(j<n){
                res.push_back(B[j]);
                j++;
            }
             
        }
        int len=res.size();
        if(len%2==0){ //偶数个
            return (double)(res[len/2-1]+res[len/2])/2;
        }
        else{
            return (double)(res[len/2]);
        }
    }
};

解法2：其实并不需要将全部数组排序，只需要将一半的数组排序即可，因为只需要中间值，这样可节省一半时间：

class Solution {
public:
    double findMedianSortedArrays(int A[], int m, int B[], int n) {
        vector<int> res;
        int i=0;
        int j=0;
        int mid=(m+n)/2+1;
         
        if(m==0){
            if(n%2==0)
                return (double)(B[n/2]+B[n/2-1])/2;
            else
                return (double)(B[n/2]);
        }
        if(n==0){
            if(m%2==0)
                return (double)(A[m/2]+A[m/2-1])/2;
            else
                return (double)(A[m/2]);
        }
        while(i<m && j<n){
                if(A[i]>=B[j]){
                    res.push_back(B[j]);
                    j++;
                }
                else{
                    res.push_back(A[i]);
                    i++;
                }
            if(res.size()==mid)
                break;
        }
        while(res.size()!=mid){
            if(i<m){
                res.push_back(A[i]);
                i++;
            }
            if(j<n){
                res.push_back(B[j]);
                j++;
            }
        }
         
        if((n+m)%2==0){
            return (res[mid-2]+res[mid-1])/2.0;
        }
        else{
            return (res[mid-1]);
        }
    }
};