1424. diagonal traversal II

1424. Diagonal traversal II

Here's a list nums , Each element is a list of integers . Please follow the rules of the following pictures , Go back to... In order nums Integer on diagonal in .

Example 1：

Input ：nums = [[1,2,3],[4,5,6],[7,8,9]]
Output ：[1,4,2,7,5,3,8,6,9]

Example 3：

Input ：nums = [[1,2,3],[4],[5,6,7],[8],[9,10,11]]
Output ：[1,4,2,5,3,8,6,9,7,10,11]

Example 4：

Input ：nums = [[1,2,3,4,5,6]]
Output ：[1,2,3,4,5,6]
The solution code is as follows ：

/** * Note: The returned array must be malloced, assume caller calls free(). */
int* findDiagonalOrder(int** nums, int numsSize, int* numsColSize, int* returnSize){
    
    int i;
     int n=numsSize;
 
  int colmax=0;
     for(i=0;i<numsSize;i++){
    
         if(numsColSize[i]>colmax){
    
             colmax=numsColSize[i];
         }
     }
       printf("n %d %d",n,colmax);
    
      
    int  *re=(int *)malloc(sizeof(int)*(n*colmax+5));
   
    int size=0;
  
    for(i=0;i<n;i++){
    
        re[size++]=nums[i][0];
        int x=0;
        int y=i;

        while(x+1<=i&&y-1>=0){
    
            
            x=x+1;
            y=y-1;
            if(numsColSize[y]>x){
    
                 re[size++]=nums[y][x];

            }

        }
        
    }
 
     for(i=1;i<colmax;i++){
    
 
         if(numsColSize[n-1]>i)
        re[size++]=nums[n-1][i];
       // printf("df");
        int x=i;
        int y=n-1;
  
        while(x+1<=colmax-1&&y-1>=0){
    
       // printf("%d %d",x+1,y-1);
            x=x+1;
            y=y-1;
            if(numsColSize[y]>x){
    
                
                 re[size++]=nums[y][x];

            }

        }
        
    }
    *returnSize=size;
    return re;

}