题意
题解
方法一：暴力法
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.size() == 0 || matrix[0].size() == 0) {
return 0;
}
int maxSide = 0;
int rows = matrix.size(), columns = matrix[0].size();
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
if (matrix[i][j] == '1') {
// 遇到一个 1 作为正方形的左上角
maxSide = max(maxSide, 1);
// 计算可能的最大正方形边长
int currentMaxSide = min(rows - i, columns - j);
for (int k = 1; k < currentMaxSide; k++) {
// 判断新增的一行一列是否均为 1
bool flag = true;
if (matrix[i + k][j + k] == '0') {
break;
}
for (int m = 0; m < k; m++) {
if (matrix[i + k][j + m] == '0' || matrix[i + m][j + k] == '0') {
flag = false;
break;
}
}
if (flag) {
maxSide = max(maxSide, k + 1);
} else {
break;
}
}
}
}
}
int maxSquare = maxSide * maxSide;
return maxSquare;
}
};
方法二：动态规划
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.size() == 0 || matrix[0].size() == 0) {
return 0;
}
int maxSide = 0;
int rows = matrix.size(), columns = matrix[0].size();
vector<vector<int>> dp(rows, vector<int>(columns));
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
if (matrix[i][j] == '1') {
if (i == 0 || j == 0) {
dp[i][j] = 1;
} else {
dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
}
maxSide = max(maxSide, dp[i][j]);
}
}
}
int maxSquare = maxSide * maxSide;
return maxSquare;
}
};
另一种解法，类似85.最大矩形
class Solution {
public int maximalSquare(char[][] matrix) {
int row = matrix.length;
if(row == 0) return 0;
int column = matrix[0].length;
int maxSide = 0;
int[][] dp = new int[row][column];
for(int i = 0; i < row; i++)
for(int j = 0; j < column; j++)
if(i == 0)
dp[i][j] = matrix[i][j] == '1' ? 1 : 0;
else if(matrix[i][j] == '1')
dp[i][j] = dp[i-1][j] + 1;
for(int i = 0; i < row; i++)
for(int j = 0; j < column; j++)
{
if(dp[i][j] == 0) continue;
int curHeight = dp[i][j];
for(int k = j; k >= 0; k--)
{
if(dp[i][k] == 0) break;
curHeight = Math.min(curHeight, dp[i][k]);
maxSide = Math.max(maxSide, Math.min(curHeight, j - k + 1));
}
}
return maxSide * maxSide;
}
}