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feat: add solutions to lc problem: No.1504
No.1504.Count Submatrices With All Ones
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‎solution/1500-1599/1504.Count Submatrices With All Ones/README.md

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@@ -58,22 +58,136 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举 + 前缀和**
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我们可以枚举矩阵的右下角 $(i, j)$，然后向上枚举矩阵的第一行 $k$，那么每一行以 $(i, j)$ 为右下角的矩阵的宽度就是 $\min_{k \leq i} \textit{g}[k][j]$，其中 $\textit{g}[k][j]$ 表示第 $k$ 行以 $(k, j)$ 为右下角的矩阵的宽度。
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因此，我们可以预处理得到二维数组 $g[i][j]$，其中 $g[i][j]$ 表示第 $i$ 行中，从第 $j$ 列向左连续的 $1$ 的个数。
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时间复杂度 $O(m^2 \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。
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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 numSubmat(self, mat: List[List[int]]) -> int:
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m, n = len(mat), len(mat[0])
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g = [[0] * n for _ in range(m)]
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for i in range(m):
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for j in range(n):
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if mat[i][j]:
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g[i][j] = 1 if j == 0 else 1 + g[i][j - 1]
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ans = 0
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for i in range(m):
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for j in range(n):
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col = inf
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for k in range(i, -1, -1):
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col = min(col, g[k][j])
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ans += col
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return ans
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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 numSubmat(int[][] mat) {
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int m = mat.length, n = mat[0].length;
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int[][] g = new int[m][n];
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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 (mat[i][j] == 1) {
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g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
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}
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}
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}
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int ans = 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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int col = 1 << 30;
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for (int k = i; k >= 0 && col > 0; --k) {
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col = Math.min(col, g[k][j]);
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ans += col;
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}
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}
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}
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return ans;
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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 numSubmat(vector<vector<int>>& mat) {
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int m = mat.size(), n = mat[0].size();
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vector<vector<int>> g(m, vector<int>(n));
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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 (mat[i][j] == 1) {
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g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
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}
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}
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}
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int ans = 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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int col = 1 << 30;
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for (int k = i; k >= 0 && col > 0; --k) {
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col = min(col, g[k][j]);
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ans += col;
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}
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}
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}
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return ans;
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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 numSubmat(mat [][]int) (ans int) {
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m, n := len(mat), len(mat[0])
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g := make([][]int, m)
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for i := range g {
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g[i] = make([]int, n)
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for j := range g[i] {
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if mat[i][j] == 1 {
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if j == 0 {
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g[i][j] = 1
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} else {
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g[i][j] = 1 + g[i][j-1]
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}
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}
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}
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}
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for i := range g {
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for j := range g[i] {
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col := 1 << 30
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for k := i; k >= 0 && col > 0; k-- {
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col = min(col, g[k][j])
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ans += col
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}
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}
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}
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return
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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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### **...**

‎solution/1500-1599/1504.Count Submatrices With All Ones/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -52,13 +52,119 @@ Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.
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### **Python3**
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```python
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class Solution:
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def numSubmat(self, mat: List[List[int]]) -> int:
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m, n = len(mat), len(mat[0])
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g = [[0] * n for _ in range(m)]
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for i in range(m):
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for j in range(n):
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if mat[i][j]:
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g[i][j] = 1 if j == 0 else 1 + g[i][j - 1]
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ans = 0
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for i in range(m):
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for j in range(n):
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col = inf
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for k in range(i, -1, -1):
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col = min(col, g[k][j])
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ans += col
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int numSubmat(int[][] mat) {
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int m = mat.length, n = mat[0].length;
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int[][] g = new int[m][n];
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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 (mat[i][j] == 1) {
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g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
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}
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}
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}
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int ans = 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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int col = 1 << 30;
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for (int k = i; k >= 0 && col > 0; --k) {
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col = Math.min(col, g[k][j]);
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ans += col;
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}
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}
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}
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return ans;
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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 numSubmat(vector<vector<int>>& mat) {
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int m = mat.size(), n = mat[0].size();
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vector<vector<int>> g(m, vector<int>(n));
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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 (mat[i][j] == 1) {
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g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
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}
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}
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}
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int ans = 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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int col = 1 << 30;
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for (int k = i; k >= 0 && col > 0; --k) {
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col = min(col, g[k][j]);
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ans += col;
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}
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}
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}
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return ans;
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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 numSubmat(mat [][]int) (ans int) {
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m, n := len(mat), len(mat[0])
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g := make([][]int, m)
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for i := range g {
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g[i] = make([]int, n)
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for j := range g[i] {
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if mat[i][j] == 1 {
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if j == 0 {
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g[i][j] = 1
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} else {
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g[i][j] = 1 + g[i][j-1]
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}
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}
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}
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}
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for i := range g {
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for j := range g[i] {
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col := 1 << 30
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for k := i; k >= 0 && col > 0; k-- {
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col = min(col, g[k][j])
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ans += col
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}
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}
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}
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return
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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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### **...**

‎solution/1500-1599/1504.Count Submatrices With All Ones/Solution.cpp

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class Solution {
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public:
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int numSubmat(vector<vector<int>>& mat) {
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int m = mat.size(), n = mat[0].size();
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vector<vector<int>> g(m, vector<int>(n));
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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 (mat[i][j] == 1) {
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g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
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}
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}
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}
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int ans = 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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int col = 1 << 30;
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for (int k = i; k >= 0 && col > 0; --k) {
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col = min(col, g[k][j]);
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ans += col;
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}
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}
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}
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return ans;
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}
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};

‎solution/1500-1599/1504.Count Submatrices With All Ones/Solution.go

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func numSubmat(mat [][]int) (ans int) {
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m, n := len(mat), len(mat[0])
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g := make([][]int, m)
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for i := range g {
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g[i] = make([]int, n)
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for j := range g[i] {
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if mat[i][j] == 1 {
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if j == 0 {
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g[i][j] = 1
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} else {
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g[i][j] = 1 + g[i][j-1]
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}
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}
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}
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}
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for i := range g {
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for j := range g[i] {
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col := 1 << 30
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for k := i; k >= 0 && col > 0; k-- {
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col = min(col, g[k][j])
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ans += col
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}
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}
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}
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return
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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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}

‎solution/1500-1599/1504.Count Submatrices With All Ones/Solution.java

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class Solution {
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public int numSubmat(int[][] mat) {
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int m = mat.length, n = mat[0].length;
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int[][] g = new int[m][n];
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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 (mat[i][j] == 1) {
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g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
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}
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}
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}
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int ans = 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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int col = 1 << 30;
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for (int k = i; k >= 0 && col > 0; --k) {
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col = Math.min(col, g[k][j]);
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ans += col;
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}
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}
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}
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return ans;
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}
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}

‎solution/1500-1599/1504.Count Submatrices With All Ones/Solution.py

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class Solution:
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def numSubmat(self, mat: List[List[int]]) -> int:
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m, n = len(mat), len(mat[0])
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g = [[0] * n for _ in range(m)]
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for i in range(m):
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for j in range(n):
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if mat[i][j]:
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g[i][j] = 1 if j == 0 else 1 + g[i][j - 1]
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ans = 0
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for i in range(m):
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for j in range(n):
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col = inf
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for k in range(i, -1, -1):
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col = min(col, g[k][j])
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ans += col
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return ans

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