LeetCode：221. 最大正方形

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’

---

解题思路

1.首先定义一个数组dp表示以[i][j]为正方形右下角的最大边长，定义一个变量max_len表示最大正方形的边长
2.状态转移方程：dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1;（取三者的最小值是因为要构成正方形）
3.注意：若左边和上边的点为’1’，则dp[i][j] = 1

代码

/** * @param {character[][]} matrix * @return {number} */
var maximalSquare = function(matrix) {
    
    let max_len = 0;
    let dp = Array.from(Array(matrix.length), () => Array(matrix[0].length).fill(0));
    for (let i = 0; i < matrix.length; i++) {
    
        for (let j = 0; j < matrix[i].length; j++) {
    
            if (matrix[i][j] === '1') {
    
                if (i === 0 || j === 0) {
    
                    dp[i][j] = 1;
                }else {
    
                    dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1;
                }
                max_len = Math.max(max_len, dp[i][j]);
            }
        }
    }
    return max_len ** 2;
};