[leetcode ]221. Maximal Square c语言

题目：

Given a 2D binary matrix filled with 0’s and 1’s, find the largest square containing all 1’s and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

题目大意：
给一个只包含0和1的矩阵，求里面值都为1的一个最大方阵，返回方阵面积

思路：参考了这里
以递推的方式求出矩阵每个点为方阵右下角时，该方阵的层数
递推公式：
如果当前点为 0，残缺的方阵，层数为0， level(i,j) = 0
如果当前点为 1，层数等于左、上、左上方阵最小的一个层数加1，level(i, j) = min(level(i-1,j), level(i,j-1), level(i-1,j-1)) + 1
代码

#define min(a, b) ((a) < (b) ? (a) : (b))
#define level(i, j) level[(i) * matrixColSize + (j)]    //方便代码阅读，一维数组转二维数组int maximalSquare(char** matrix, int matrixRowSize, int matrixColSize) {int *level = NULL;  /* 矩阵上每个点为方阵右下角时方阵的层数 */int i, j;int maxLevel = 0;  if(!matrix || !matrixRowSize || !matrixColSize)return 0;level = (int *)malloc(matrixRowSize * matrixColSize * sizeof(int));if(!level)return 0;/* 初始化第一行和第一列为方阵右下角时的层数 */for(j = 0; j < matrixColSize; j++){level(0, j) = matrix[0][j] - '0';if(maxLevel < level(0, j))maxLevel = level(0, j);}for(i = 0; i < matrixRowSize; i++){level(i, 0) = matrix[i][0] - '0';if(maxLevel < level(i, 0))maxLevel = level(i, 0);}/* 遍历矩阵 */for(i = 1; i < matrixRowSize; i++){for(j = 1; j < matrixColSize; j++){if(matrix[i][j] == '0')level(i, j) = 0;elselevel(i, j) = min(min(level(i-1, j), level(i, j-1)), level(i-1, j-1)) + 1;if(maxLevel < level(i, j))maxLevel = level(i, j);}}/* for debugfor(i = 0; i < matrixRowSize; i++){for(j = 0; j < matrixColSize; j++){printf("%c ", matrix[i][j]);}printf("\n");}printf("\n");for(i = 0; i < matrixRowSize; i++){for(j = 0; j < matrixColSize; j++){printf("%d ", level(i, j));}printf("\n");}printf("\n");
*/free(level);return (maxLevel * maxLevel);
}

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