LeetCode 2.Median of Two Sorted Arrays

Median of Two Sorted Arrays

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)).

Solution

#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:

    double findMedianSortedArrays(int A[], int m, int B[], int n) 
    {
        int  findNum=0;
        if((m+n)%2==0)
            findNum=2;
        else
            findNum=1;

        int leftOffsetA=0,leftOffsetB=0;
        int rightOffsetA=0,rightOffsetB=0;
        while(leftOffsetA+leftOffsetB+rightOffsetA+rightOffsetB!=m+n-findNum)
        {
            
            if(leftOffsetA==m|| rightOffsetA==m)
            {
                //左右端都在B中
                int halfLenthOfPartB=(n-1-leftOffsetB-rightOffsetB)/2;
                leftOffsetB+=halfLenthOfPartB;
                rightOffsetB+=halfLenthOfPartB;
                break;
            }
            else if(leftOffsetB==n|| rightOffsetB==n)
            {
                //左右端都在A中
                int halfLenthOfPartA=(m-1-leftOffsetA-rightOffsetA)/2;
                leftOffsetA+=halfLenthOfPartA;
                rightOffsetA+=halfLenthOfPartA;
                break;
            }
            else
            {
                //左端还分别在A、B中
                if(A[leftOffsetA]<B[leftOffsetB])
                    leftOffsetA++;
                else
                    leftOffsetB++;

                if(A[m-1-rightOffsetA]>B[n-1-rightOffsetB])
                    rightOffsetA++;
                else
                    rightOffsetB++;
            }
        }
    //    cout<<"leftOffsetA:"<<leftOffsetA<<" rightOffsetA"<<rightOffsetA<<endl;
    //    cout<<"leftOffsetB:"<<leftOffsetB<<" rightOffsetB"<<rightOffsetB<<endl;
        
        //取出两段截断的int数据
        vector<int> ivec;
        for(int index_i=leftOffsetA;index_i<=m-1-rightOffsetA &&index_i<m;index_i++)
        {
            ivec.push_back(A[index_i]);
            //cout<<"A中取出:"<<A[index_i]<<endl;
        }
        for(int index_i=leftOffsetB;index_i<=n-1-rightOffsetB &&index_i<n;index_i++)
        {
            ivec.push_back(B[index_i]);
            //cout<<"B中取出:"<<B[index_i]<<endl;
        }
        if(findNum==1)
            return ivec[0];
        else
            return (ivec[0]+ivec[1])/2.0f;
        
    }
};