Daily question 1 · 1260. Two dimensional network migration · simulation

link ：https://leetcode.cn/problems/shift-2d-grid/solution/chun-c-by-xun-ge-v-7ky6/
source ： Power button （LeetCode）
The copyright belongs to the author . Commercial reprint please contact the author for authorization , Non-commercial reprint please indicate the source . 

subject

 

Example

 

Ideas

The topic has been told to us every time 「 transfer 」 The operation will trigger the following activities ：

• be located grid[i][j]  The element will be moved to  grid[i][j + 1].
• be located  grid[i][n - 1] The element will be moved to  grid[i + 1][0].
• be located grid[m - 1][n - 1]  The element will be moved to  grid[0][0].

Directly simulate according to the given requirements of the topic

Little details ： Use the remainder to merge the third special case into the second

Code

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** shiftGrid(int** grid, int gridSize, int* gridColSize, int k, int* returnSize, int** returnColumnSizes){
    int ** res = malloc(sizeof(int *) * gridSize);
    *returnColumnSizes = malloc(sizeof(int) * gridSize);
    *returnSize = gridSize;
    for(int i = 0; i < gridSize; i++)// Initialize variable 
    {
        res[i] = malloc(sizeof(int) * gridColSize[0]);
        (*returnColumnSizes)[i] = gridColSize[0];
        for(int j = 0; j < gridColSize[0]; j++)// take grid Array to res
        {
            res[i][j] = grid[i][j];
        }
    }
    while(k)// Direct simulation 
    {
        for(int i = 0; i < gridSize; i++)// Brute force loop traversal array 
        {
            for(int j = 0; j < gridColSize[0]; j++)
            {
                if(j == gridColSize[0]-1)// First deal with special situations 
                {
                    res[((i+1)%gridSize)][0] = grid[i][j];// Use the remainder to merge the third special case into the second 
                }
                else 
                {
                    res[i][j+1] = grid[i][j];
                }
            }
        }
        for(int i = 0; i < gridSize; i++)// Change the original array 
        {
            for(int j = 0; j < gridColSize[0]; j++)
            {
                grid[i][j] = res[i][j];
            }
        }
        k--;
    }
    return res;
}

 author ：xun-ge-v
 link ：https://leetcode.cn/problems/shift-2d-grid/solution/chun-c-by-xun-ge-v-7ky6/
 source ： Power button （LeetCode）
 The copyright belongs to the author . Commercial reprint please contact the author for authorization , Non-commercial reprint please indicate the source .

Time and space complexity