LEET code--Median of Two Sorted Arrays

LEET code--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)).

一開始直觀  就是先檢查有沒有一個數列完全比另外一個大，但是這樣做會讓CODE變長LOL

接著在去慢慢比較找中位數

這是比較笨的寫法，看到有人用更神的
還要再學習阿~

int checkoverlap(int* nums1, int nums1Size, int* nums2, int nums2Size){
    int flag=0;
    
    if(nums1[ nums1Size-1 ] < nums2[0] ){//nums1 is less than 2
        return 1;
    }else if(nums2[ nums2Size-1 ] < nums1[0]){//nums2 is less than 1
        return 2;
    }else {// overlap
        return 3;
    }
    
}
double case1(int* nums1, int nums1Size, int* nums2, int nums2Size){
    int medin=(nums1Size+nums2Size)/2 +1 ;//because the devide won't round up   5/2=2,6/2=3
    
    if( nums1Size < medin ){//inside the nums2
        if( (nums1Size+nums2Size)%2==0 ){
            return ( (medin-nums1Size ==1) ? (nums1[nums1Size-1]+nums2[0])/2.0 : (nums2[medin - nums1Size-1]+nums2[medin - nums1Size-2] )/2.0 );
                    //medin is in the first nums2: not in the first
        }else {
            return nums2[medin - nums1Size -1 ];
        }
                
    }else if( nums1Size > medin ){//inside the nums1
        return ( ((nums1Size+nums2Size)%2==0 ) ?  (nums1[medin -1]+nums1[medin -2] )/2.0 : nums1[medin  -1 ]  );
                //even: odd
    }else if( nums1Size == medin ){//medin is in the last number of nums1
        return ( (nums1Size+nums2Size)%2==0 ?  (nums1[nums1Size-1] + nums1[nums1Size-2]  )/2.0 : nums1[nums1Size-1] );
                //even, last2 number in nums1: last number in nums1
    }
}

double case2(int* nums1, int nums1Size, int* nums2, int nums2Size){
    int medin=(nums1Size+nums2Size)/2 +1 ;//because the devide won't round up   5/2=2,6/2=3
    if( nums2Size < medin ){//inside the nums1
        if( (nums1Size+nums2Size)%2==0 ){
            return ( (medin-nums2Size ==1) ? (nums2[nums2Size-1]+nums1[0])/2.0 : (nums1[medin - nums2Size-1]+nums1[medin - nums2Size-2] )/2.0 );
                    //medin is in the first nums2: not in the first
        }else {
            return nums1[medin - nums2Size -1 ];
        }
                
    }else if( nums2Size > medin ){//inside the nums2
        return ( ((nums1Size+nums2Size)%2==0 ) ?  (nums2[medin -1]+nums2[medin -2] )/2.0 : nums2[medin  -1 ]  );
                //even: odd
    }else if( nums2Size == medin ){//medin is in the last number of nums1
        return ( (nums1Size+nums2Size)%2==0 ?  (nums2[nums2Size-1] + nums2[nums2Size-2]  )/2.0 : nums2[nums2Size-1] );
                //even, last2 number in nums1: last number in nums1
    }
}


double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    if(nums1Size==0){
         if(nums2Size ==1)return nums2[0];
        return ( nums2Size%2==0 ? (nums2[nums2Size/2]+nums2[(nums2Size/2) - 1])/2.0  : nums2[nums2Size/2] );//even:odd
    }
    if(nums2Size==0){
        if(nums1Size ==1)return nums1[0];
        return ( nums1Size%2==0 ? (nums1[nums1Size/2]+nums1[nums1Size/2 - 1])/2.0  : nums1[nums1Size/2] );
    }
    //find the overlap
    int medin=(nums1Size+nums2Size)/2 +1 ;//because the devide won't round up   5/2=2,6/2=3
    int sum= nums1Size + nums2Size;
    int nums1P,nums2P,remin=0;
    int previous=0,current=0;
    
    switch ( checkoverlap( nums1,  nums1Size,  nums2,  nums2Size) ){
        case 1://nums1 < nums2
            return case1( nums1,  nums1Size,  nums2,  nums2Size);
        
            break;
        case 2://nums2 < nums1
            return case2( nums1,  nums1Size,  nums2,  nums2Size);
            break;
        case 3://overlap nums1  nums2
            
            nums1P=nums2P=1;
            //count how many number need to process
            remin=medin;
            while ( nums2P<=nums2Size || nums1P<=nums1Size ){
                if( nums1P<= nums1Size  && nums1[nums1P-1] <= nums2[nums2P -1] ){
                    previous = current;
                    current = nums1[nums1P-1];
                    nums1P++;
                    remin--;
                    
                    
                    //if( nums1P > nums1Size)remin--;
                }else if(  nums2P<=nums2Size && nums2[nums2P -1 ] < nums1[nums1P - 1] ){
                    previous = current ;
                    current = nums2[nums2P-1];
                    nums2P++;
                    remin--;
                    
                    
                    //if( nums2P > nums2Size)remin--;
                }else if( nums2P > nums2Size ){
                    previous = current;
                    current = nums1[nums1P-1];
                    nums1P++;
                    remin--;

                }else if( nums1P > nums1Size){
                    previous = current ;
                    current = nums2[nums2P-1];
                    nums2P++;
                    remin--;
                }
                if(remin<=0 ){
                    return ( ((nums1Size+nums2Size)%2==0 ) ?  (previous+ current)/2.0 :current);
                    break;
                }
            }
            
            
            break;
    }
    
}