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feat: add solutions to lc problem: No.0221. Maximal Square
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‎solution/0200-0299/0221.Maximal Square/README.md

+114-2
Original file line numberDiff line numberDiff line change
@@ -42,27 +42,139 @@
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<li><code>matrix[i][j]</code> 为 <code>'0'</code> 或 <code>'1'</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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动态规划。
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设 `dp[i + 1][j + 1]` 表示以下标 `(i, j)` 作为正方形右下角的最大正方形边长。
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当 `matrix[i][j] == '1'`, `dp[i + 1][j + 1] = min(dp[i][j + 1], dp[i + 1][j], dp[i][j]) + 1`，否则 `dp[i + 1][j + 1] = 0`。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maximalSquare(self, matrix: List[List[str]]) -> int:
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m, n = len(matrix), len(matrix[0])
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dp = [[0] * (n + 1) for _ in range(m + 1)]
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mx = 0
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for i in range(m):
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for j in range(n):
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if matrix[i][j] == '1':
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dp[i + 1][j + 1] = min(dp[i][j + 1], dp[i + 1][j], dp[i][j]) + 1
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mx = max(mx, dp[i + 1][j + 1])
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return mx * mx
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maximalSquare(char[][] matrix) {
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int m = matrix.length, n = matrix[0].length;
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int[][] dp = new int[m + 1][n + 1];
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int mx = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (matrix[i][j] == '1') {
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dp[i + 1][j + 1] = Math.min(Math.min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
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mx = Math.max(mx, dp[i + 1][j + 1]);
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}
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}
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}
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return mx * mx;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maximalSquare(vector<vector<char>>& matrix) {
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int m = matrix.size(), n = matrix[0].size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
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int mx = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (matrix[i][j] == '1') {
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dp[i + 1][j + 1] = min(min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
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mx = max(mx, dp[i + 1][j + 1]);
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}
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}
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}
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return mx * mx;
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}
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};
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```
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### **Go**
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```go
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func maximalSquare(matrix [][]byte) int {
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m, n := len(matrix), len(matrix[0])
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dp := make([][]int, m+1)
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for i := 0; i <= m; i++ {
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dp[i] = make([]int, n+1)
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}
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mx := 0
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for i := 0; i < m; i++ {
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for j := 0; j < n; j++ {
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if matrix[i][j] == '1' {
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dp[i+1][j+1] = min(min(dp[i][j+1], dp[i+1][j]), dp[i][j]) + 1
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mx = max(mx, dp[i+1][j+1])
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}
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}
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}
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return mx * mx
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **C#**
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```cs
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public class Solution {
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public int MaximalSquare(char[][] matrix) {
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int m = matrix.Length, n = matrix[0].Length;
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var dp = new int[m + 1, n + 1];
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int mx = 0;
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for (int i = 0; i < m; ++i)
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{
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for (int j = 0; j < n; ++j)
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{
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if (matrix[i][j] == '1')
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{
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dp[i + 1, j + 1] = Math.Min(Math.Min(dp[i, j + 1], dp[i + 1, j]), dp[i, j]) + 1;
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mx = Math.Max(mx, dp[i + 1, j + 1]);
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}
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}
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}
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return mx * mx;
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}
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}
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```
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### **...**

‎solution/0200-0299/0221.Maximal Square/README_EN.md

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@@ -46,13 +46,120 @@
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### **Python3**
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```python
49-
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class Solution:
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def maximalSquare(self, matrix: List[List[str]]) -> int:
51+
m, n = len(matrix), len(matrix[0])
52+
dp = [[0] * (n + 1) for _ in range(m + 1)]
53+
mx = 0
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for i in range(m):
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for j in range(n):
56+
if matrix[i][j] == '1':
57+
dp[i + 1][j + 1] = min(dp[i][j + 1], dp[i + 1][j], dp[i][j]) + 1
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mx = max(mx, dp[i + 1][j + 1])
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return mx * mx
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```
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### **Java**
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```java
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class Solution {
66+
public int maximalSquare(char[][] matrix) {
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int m = matrix.length, n = matrix[0].length;
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int[][] dp = new int[m + 1][n + 1];
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int mx = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (matrix[i][j] == '1') {
73+
dp[i + 1][j + 1] = Math.min(Math.min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
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mx = Math.max(mx, dp[i + 1][j + 1]);
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}
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}
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}
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return mx * mx;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
88+
int maximalSquare(vector<vector<char>>& matrix) {
89+
int m = matrix.size(), n = matrix[0].size();
90+
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
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int mx = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
94+
if (matrix[i][j] == '1') {
95+
dp[i + 1][j + 1] = min(min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
96+
mx = max(mx, dp[i + 1][j + 1]);
97+
}
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}
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}
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return mx * mx;
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}
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};
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```
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### **Go**
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```go
108+
func maximalSquare(matrix [][]byte) int {
109+
m, n := len(matrix), len(matrix[0])
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dp := make([][]int, m+1)
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for i := 0; i <= m; i++ {
112+
dp[i] = make([]int, n+1)
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}
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mx := 0
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for i := 0; i < m; i++ {
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for j := 0; j < n; j++ {
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if matrix[i][j] == '1' {
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dp[i+1][j+1] = min(min(dp[i][j+1], dp[i+1][j]), dp[i][j]) + 1
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mx = max(mx, dp[i+1][j+1])
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}
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}
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}
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return mx * mx
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}
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func max(a, b int) int {
127+
if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
134+
if a < b {
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return a
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}
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return b
138+
}
139+
```
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141+
### **C#**
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```cs
144+
public class Solution {
145+
public int MaximalSquare(char[][] matrix) {
146+
int m = matrix.Length, n = matrix[0].Length;
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var dp = new int[m + 1, n + 1];
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int mx = 0;
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for (int i = 0; i < m; ++i)
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{
151+
for (int j = 0; j < n; ++j)
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{
153+
if (matrix[i][j] == '1')
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{
155+
dp[i + 1, j + 1] = Math.Min(Math.Min(dp[i, j + 1], dp[i + 1, j]), dp[i, j]) + 1;
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mx = Math.Max(mx, dp[i + 1, j + 1]);
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}
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}
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}
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return mx * mx;
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}
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}
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```
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### **...**

‎solution/0200-0299/0221.Maximal Square/Solution.cpp

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@@ -0,0 +1,17 @@
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class Solution {
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public:
3+
int maximalSquare(vector<vector<char>>& matrix) {
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int m = matrix.size(), n = matrix[0].size();
5+
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
6+
int mx = 0;
7+
for (int i = 0; i < m; ++i) {
8+
for (int j = 0; j < n; ++j) {
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if (matrix[i][j] == '1') {
10+
dp[i + 1][j + 1] = min(min(dp[i][j + 1], dp[i + 1][j]), dp[i][j]) + 1;
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mx = max(mx, dp[i + 1][j + 1]);
12+
}
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}
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}
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return mx * mx;
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}
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};

‎solution/0200-0299/0221.Maximal Square/Solution.cs

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Original file line numberDiff line numberDiff line change
@@ -1,40 +1,19 @@
1-
// https://leetcode.com/problems/maximal-square/
2-
3-
using System;
4-
5-
public partial class Solution
6-
{
7-
public int MaximalSquare(char[][] matrix)
8-
{
9-
var lengthI = matrix.Length;
10-
var lengthJ = lengthI == 0 ? 0 : matrix[0].Length;
11-
if (lengthI == 0 || lengthJ == 0) return 0;
12-
13-
var lenI = new int[lengthI, lengthJ];
14-
var lenJ = new int[lengthI, lengthJ];
15-
var f = new int[lengthI, lengthJ];
16-
var maxSideLength = 0;
17-
for (var i = 0; i < lengthI; ++i)
1+
public class Solution {
2+
public int MaximalSquare(char[][] matrix) {
3+
int m = matrix.Length, n = matrix[0].Length;
4+
var dp = new int[m + 1, n + 1];
5+
int mx = 0;
6+
for (int i = 0; i < m; ++i)
187
{
19-
for (var j = 0; j < lengthJ ; ++j)
8+
for (int j = 0; j < n; ++j)
209
{
2110
if (matrix[i][j] == '1')
2211
{
23-
lenI[i, j] = i == 0 ? 1 : lenI[i - 1, j] + 1;
24-
lenJ[i, j] = j == 0 ? 1 : lenJ[i, j - 1] + 1;
25-
if (i == 0 || j == 0)
26-
{
27-
f[i, j] = 1;
28-
}
29-
else
30-
{
31-
f[i, j] = Math.Min(Math.Min(f[i - 1, j - 1] + 1, lenI[i, j]), lenJ[i, j]);
32-
}
33-
if (f[i, j] > maxSideLength) maxSideLength = f[i, j];
12+
dp[i + 1, j + 1] = Math.Min(Math.Min(dp[i, j + 1], dp[i + 1, j]), dp[i, j]) + 1;
13+
mx = Math.Max(mx, dp[i + 1, j + 1]);
3414
}
3515
}
3616
}
37-
38-
return maxSideLength * maxSideLength;
17+
return mx * mx;
3918
}
4019
}

‎solution/0200-0299/0221.Maximal Square/Solution.go

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1+
func maximalSquare(matrix [][]byte) int {
2+
m, n := len(matrix), len(matrix[0])
3+
dp := make([][]int, m+1)
4+
for i := 0; i <= m; i++ {
5+
dp[i] = make([]int, n+1)
6+
}
7+
mx := 0
8+
for i := 0; i < m; i++ {
9+
for j := 0; j < n; j++ {
10+
if matrix[i][j] == '1' {
11+
dp[i+1][j+1] = min(min(dp[i][j+1], dp[i+1][j]), dp[i][j]) + 1
12+
mx = max(mx, dp[i+1][j+1])
13+
}
14+
}
15+
}
16+
return mx * mx
17+
}
18+
19+
func max(a, b int) int {
20+
if a > b {
21+
return a
22+
}
23+
return b
24+
}
25+
26+
func min(a, b int) int {
27+
if a < b {
28+
return a
29+
}
30+
return b
31+
}

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