【C语言入门】第十四天

【例题1】1260. 二维网格迁移 - 力扣（LeetCode）

/**
 * 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){
    // 创建数组
    int** p = (int**)malloc(sizeof(int*)*gridSize);
    for(int i = 0;i<gridSize;i++){
        p[i] = (int*)malloc(sizeof(int)*(*gridColSize));
    }
    //实现逻辑
    int i,j;
    for(i=0;i<gridSize;i++){
        for(j=0;j<*gridColSize;j++){
            int value = ((i*(*gridColSize))+j+k)%(gridSize*(*gridColSize));
            p[value/(*gridColSize)][value%(gridSize)] = grid[i][j];
        }
    }
    //确定返回参数
    *returnSize = gridSize;
    *returnColumnSizes = (int*)malloc(sizeof(int)*gridSize);
    for(i=0;i<gridSize;i++){
        (*returnColumnSizes)[i] = *gridColSize;
    }
    return p;
}

【例题2】661. 图片平滑器 - 力扣（LeetCode）

/**
 * 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** imageSmoother(int** img, int imgSize, int* imgColSize, int* returnSize, int** returnColumnSizes){
    int i,j;
    // 创建数组
    int** p = (int**)malloc(sizeof(int*)*imgSize);
    for(i=0;i<imgSize;i++){
        p[i] = (int*)malloc(sizeof(int)*(*imgColSize));
    }
    //实现逻辑
    for(i=0;i<imgSize;i++){
        for(j=0;j<*imgColSize;j++){
            // 核心代码
            int sum=0;int fenmu = 0;
            for(int m=i-1;m<=i+1;m++){
                for(int n=j-1;n<=j+1;n++){
                    if(m >=0 && m < imgSize && n>=0 && n<(*imgColSize)){
                        sum+=img[m][n];fenmu++;
                    }
                }
            }
            p[i][j] = (int)(sum*1.0/fenmu);
        }
    }
    //规定返回参数
    *returnSize = imgSize;
    *returnColumnSizes = (int*)malloc(sizeof(int)*imgSize);
    for(i=0;i<imgSize;i++){
        (*returnColumnSizes)[i] = *imgColSize;
    }
    return p;
}

【例题3】1314. 矩阵区域和 - 力扣（LeetCode）

/**
 * 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** matrixBlockSum(int** mat, int matSize, int* matColSize, int k, int* returnSize, int** returnColumnSizes){
    // 创建数组
    int** p = (int**)malloc(sizeof(int*)*matSize);
    for(int i=0;i<matSize;i++){
        p[i] = (int*)malloc(sizeof(int)*(*matColSize));
    }
    // 实现逻辑
    for(int i=0;i<matSize;i++){
        for(int j=0;j<(*matColSize);j++){
            int sum = 0;
            for(int m=i-k;m<=i+k;m++){
                for(int n=j-k;n<=j+k;n++){
                    if(m>=0 && m<matSize && n>=0 && n<(*matColSize)){
                        sum += mat[m][n];
                    }
                }
            }
            p[i][j] = sum;
        }
    }
    // 规定返回参数值
    *returnSize = matSize;
    *returnColumnSizes = (int*)malloc(sizeof(int)*matSize);
    for(int i=0;i<matSize;i++){
        (*returnColumnSizes)[i] = *matColSize;
    }
    return p;
}

【例题4】LCP 44. 开幕式焰火 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int arr[1001];
void transer(struct TreeNode* root){
    if(root == NULL) return;
    arr[root->val] = 1;
    transer(root->left);
    transer(root->right);
}
int numColor(struct TreeNode* root){
    if(root == NULL) return 0;
    memset(arr,0,sizeof(arr));
    transer(root);
    int ans = 0;
    for(int i=0;i<sizeof(arr)/sizeof(int);i++){
        if(arr[i] == 1) ans++;
    }
    return ans;
}

【例题5】397. 整数替换 - 力扣（LeetCode）

int integerReplacement(int n){
    unsigned int m = n;
    if(m==1) return 0;
    if(m%2==0) return 1+integerReplacement(m/2);
    else return 1+integerReplacement(m-1)<1+integerReplacement(m+1)?1+integerReplacement(m-1):1+integerReplacement(m+1);
}