在一个由 '0'
和 '1'
组成的二维矩阵内,找到只包含 '1'
的最大正方形,并返回其面积。
示例 1:
输入:matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出:4
示例 2:
输入:matrix = [["0","1"],["1","0"]]
输出:1
示例 3:
输入:matrix = [["0"]]
输出:0
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j]
为'0'
或'1'
/**
* @author [头号玩儿家(QtumOne)](https://qtum.one) <QtumOne@gmail.com>
* @param {character[][]} matrix
* @return {number}
*/
// dp[i, j] = min(dp[i - 1, j], dp[i, j - 1], dp[i - 1, j - 1]) + 1;
var maximalSquare = function(matrix) {
const m = matrix.length, n = matrix[0].length;
let ans = 0;
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
if (+matrix[i][j]) {
matrix[i][j] = (Math.min(matrix[i - 1]?.[j], matrix[i][j - 1], matrix[i - 1]?.[j - 1]) || 0) + 1;
ans = Math.max(ans, matrix[i][j]);
}
}
}
return ans * ans;
};