LeetCode Problem No.4

Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 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)).
You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

Solution 1:

int len1 = nums1.size();
int len2 = nums2.size();
vector list;
//每次都从nums1和nums2中找小的push_back进去，
//直到vector的size 为 (len1+len2)/2;
if((len1+len2)%2==0)
{
    return list[ (len1+len2)/2];
}
else
{
    return (list[ (len1+len2)/2]+ list[ (len1+len2)/2-1])/2.0;
}
//缺点：申请了新的内存，时间是(m+n)/2 不是long(m+n)
//但是时间效果比使用二分查找速度快

Solution 2:

//开辟新的空间，寻找中位数，实际上和上面思路基本相同
//缺点：时间复杂度不是log(m+n)
class Solution {
public:
    double findMedianSortedArrays(vector& nums1, vector& nums2) {
        int num1 = nums1.size();
        int num2 = nums2.size();
        if(num1 == 0 && num2 == 0)
            return 0.0;
        int total = num1 + num2;
        int mid = total / 2;
        int flag = total % 2;
        int left1 = 0;
        int left2 = 0;
        int count = 0;
        if(flag == 0)
        {
            int res = 0;
            while(count < mid)
            {
                if(left1nums2[left2])
                {
                    res2 = nums2[left2];
                }
                else
                {
                    res2 = nums1[left1];
                }       
            }
            else if(left1

Solution 3:

//题目中明确要求时间复杂度为log(m+n)
//搜索时应该使用二分方法进行搜索
//常见的二分法是针对一个数组设置left和right标志，进算mid位与目标值大小关系来确定left和right的新值
//数组nums1的长度为m，nums2的长度为n
//假设存在i和j,i在[0,m-1]范围内,j在[0,n-1]范围内
//能使得nums1[0]~nums1[i-1]和nums2[0]~num2[j-1]恰好在中位左侧
//nums1[i]~nums1[m-1]和nums2[j]~nums2[j-1]恰好在中位右侧
//那么 i+j = m - i + j-i 或 i+j = m-i+n-j+1
//可以得到 i+j = (m+n+1)/2
//如果 i 属于[0,m-1] 那么 j = (m+n+1)/2 - i
//如果m>n 则 j可能是负数，因此这种情况下，我们交换nums1和nums2即可
//则问题就变成了找到合适的i,j使得nums2[j-1]<=nums1[i]并且nums1[i-1]<=nums2[j],由于可以由i计算j那么在[0,m-1]中寻找i即可，可以使用二分法查找
class Solution {
public:
    double findMedianSortedArrays(vector& nums1, vector& nums2) {
        int m = nums1.size();
        int n = nums2.size();
        if(m>n)
        {
            return findMedianSortedArrays(nums2,nums1);
        }
        
        int left = 0;
        int right = m;
        int mid = (m+n+1)/2;
        while(left<=right)
        {
            int i = (left+right)/2;
            int j = mid - i;
            if(inums1[i])
            {
                left = i+1;
            }
            else if(i>left&&nums1[i-1]>nums2[j])
            {
                right = i-1;
            }
            else
            {
                int maxleft = 0;
                if(i==0)
                {
                    maxleft=nums2[j-1];
                }
                else if(j==0)
                {
                    maxleft=nums1[i-1];    
                }
                else
                {
                    if(nums2[j-1]>nums1[i-1])
                    {
                        maxleft=nums2[j-1];
                    }
                    else
                    {
                        maxleft=nums1[i-1];
                    }
                }
                
                if((m+n)%2==1)
                {
                    return maxleft;
                }
                
                int minright = 0;
                if(i==m) 
                {
                    minright=nums2[j];
                }
                else if(j==n)
                {
                    minright=nums1[i];
                }
                else
                {
                     if(nums2[j]>nums1[i])
                    {
                        minright=nums1[i];
                    }
                    else
                    {
                        minright=nums2[j];
                    }
                }
                return (maxleft+minright)/2.0;
            }
        }
        return 0.0;
    }