LeetCode 221 Maximal Square (最大子正方形 dp)

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

For example, given the following matrix:

1 0 1 0 01 0 1 1 11 1 1 1 11 0 0 1 0

Return 4.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

题目链接：https://leetcode.com/problems/maximal-square/

题目分析：考虑使用动态规划解决，先预处理出每个点向上向左最远能延伸的距离，设dp[i][j]表示以点(i,j)为右下角的正方形的最大边长，很容易得到状态转移方程：dp[i][j] = min(left[i][j - 1], up[i - 1][j], dp[i - 1][j - 1])) + 1

public class Solution {    public int maximalSquare(char[][] matrix) {        int n = matrix.length;        if (n == 0) {            return 0;        }        int m = matrix[0].length;        int[][] dp = new int[n + 5][m + 5];        int[][] up = new int[n + 5][m + 5];        int[][] left = new int[n + 5][m + 5];        boolean has = false;        for (int j = 0; j < m; j ++) {            if (matrix[0][j] == '1') {                up[0][j] = 1;                has = true;            }            dp[0][j] = up[0][j];        }        for (int i = 0; i < n; i ++) {            if (matrix[i][0] == '1') {                left[i][0] = 1;                has = true;            }            dp[i][0] = left[i][0];        }           for (int i = 0; i < n; i ++) {            for (int j = 0; j < m; j ++) {                if (i > 0) {                    up[i][j] = (matrix[i][j] == '1') ? up[i - 1][j] + 1 : 0;                }                if (j > 0) {                    left[i][j] = (matrix[i][j] == '1') ? left[i][j - 1] + 1 : 0;                }            }        }        int ans = has ? 1 : 0;        dp[0][0] = (matrix[0][0] == '1') ? 1 : 0;        for (int i = 1; i < n; i ++) {            for (int j = 1; j < m; j ++) {                if (matrix[i][j] == '1') {                    dp[i][j] = Math.min(left[i][j - 1], Math.min(up[i - 1][j], dp[i - 1][j - 1])) + 1;                    ans = Math.max(ans, dp[i][j]);                }            }        }        return ans * ans;    }}

上面的代码写得很丑。。。其实仔细观察可以发现，up和left数组是多余的，dp[i][j] = min(left[i][j - 1], up[i - 1][j], dp[i - 1][j - 1])) + 1这个状态转移方程其实等价于dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])) + 1，原来转移方程的原理是若当前点是'1'且dp[i - 1][j - 1]已经构成了一个边长为k的正方形，若点(i, j)向上向左均可以延伸k个单位长度，那么dp[i][j]可以得到一个边长为k+1的正方形，但其实满足以上三个条件时，显然可以得出dp[i - 1][j]和dp[i][j - 1]都能得到一个边长为k的正方形的结论(画个图可以很清楚的看出)，这样两个转移方程就等价了，为了方便处理，将整个矩阵向右下移动一个单位长度，所以其实这样写即可，击败了78%

public class Solution {    public int maximalSquare(char[][] matrix) {        int n = matrix.length;        if (n == 0) {            return 0;        }        int m = matrix[0].length;        int[][] dp = new int[n + 1][m + 1];        int ans = 0;        for (int i = 1; i <= n; i ++) {            for (int j = 1; j <= m; j ++) {                if (matrix[i - 1][j - 1] == '1') {                    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;                    ans = Math.max(ans, dp[i][j]);                }            }        }        return ans * ans;    }}

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