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feat: add solutions to lcof2 problem: No.040
No.040.Maximal Rectangle
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‎lcof2/剑指 Offer II 040. 矩阵中最大的矩形/README.md

+180-5
Original file line numberDiff line numberDiff line change
@@ -69,10 +69,9 @@
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**方法一：单调栈**
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72-
- 首先在柱状图中求最大矩形面积可以通过单调栈，维护每一列的左边第一个比它小的位置 $L$，和右边第一个比它小的位置 $R$，就能得到以这一列为高的最大矩形面积为 $(R-L-1)*h$。
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- 考虑每一行作为底边的柱状图中，能够得到的最大的矩形面积。再对每一行的最大面积取 $max$ 就是最终的答案。
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- 柱状图中每一列的高可以通过类似前缀和的方式去维护。
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- 假设矩阵大小为 $n*m$，那么时间复杂为 $O(nm)$，空间复杂度为 $O(m)$。
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把每一行视为柱状图的底部，对每一行求柱状图的最大面积即可。
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时间复杂度 $O(mn)$，其中 $m$ 表示 $matrix$ 的行数，$n$ 表示 $matrix$ 的列数。
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<!-- tabs:start -->
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@@ -81,19 +80,97 @@
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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 maximalRectangle(self, matrix: List[List[str]]) -> int:
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if not matrix:
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return 0
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heights = [0] * len(matrix[0])
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ans = 0
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for row in matrix:
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for j, v in enumerate(row):
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if v == "1":
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heights[j] += 1
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else:
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heights[j] = 0
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ans = max(ans, self.largestRectangleArea(heights))
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return ans
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def largestRectangleArea(self, heights: List[int]) -> int:
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n = len(heights)
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stk = []
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left = [-1] * n
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right = [n] * n
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for i, h in enumerate(heights):
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while stk and heights[stk[-1]] >= h:
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stk.pop()
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if stk:
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left[i] = stk[-1]
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stk.append(i)
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stk = []
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for i in range(n - 1, -1, -1):
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h = heights[i]
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while stk and heights[stk[-1]] >= h:
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stk.pop()
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if stk:
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right[i] = stk[-1]
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stk.append(i)
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return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))
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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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91124
```java
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class Solution {
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public int maximalRectangle(String[] matrix) {
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if (matrix == null || matrix.length == 0) {
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return 0;
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}
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int n = matrix[0].length();
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int[] heights = new int[n];
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int ans = 0;
133+
for (var row : matrix) {
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for (int j = 0; j < n; ++j) {
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if (row.charAt(j) == '1') {
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heights[j] += 1;
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} else {
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heights[j] = 0;
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}
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}
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ans = Math.max(ans, largestRectangleArea(heights));
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}
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return ans;
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}
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private int largestRectangleArea(int[] heights) {
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int res = 0, n = heights.length;
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Deque<Integer> stk = new ArrayDeque<>();
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int[] left = new int[n];
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int[] right = new int[n];
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Arrays.fill(right, n);
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for (int i = 0; i < n; ++i) {
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while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
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right[stk.pop()] = i;
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}
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left[i] = stk.isEmpty() ? -1 : stk.peek();
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stk.push(i);
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}
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for (int i = 0; i < n; ++i) {
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res = Math.max(res, heights[i] * (right[i] - left[i] - 1));
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}
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return res;
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}
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}
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```
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95167
### **C++**
96168

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- 首先在柱状图中求最大矩形面积可以通过单调栈，维护每一列的左边第一个比它小的位置 $L$，和右边第一个比它小的位置 $R$，就能得到以这一列为高的最大矩形面积为 $(R-L-1)*h$。
170+
- 考虑每一行作为底边的柱状图中，能够得到的最大的矩形面积。再对每一行的最大面积取 $max$ 就是最终的答案。
171+
- 柱状图中每一列的高可以通过类似前缀和的方式去维护。
172+
- 假设矩阵大小为 $n*m$，那么时间复杂为 $O(nm)$，空间复杂度为 $O(m)$。
173+
97174
```cpp
98175
class Solution {
99176
public:
@@ -130,6 +207,104 @@ public:
130207
};
131208
```
132209
210+
```cpp
211+
class Solution {
212+
public:
213+
int maximalRectangle(vector<string>& matrix) {
214+
if (matrix.empty()) return 0;
215+
int n = matrix[0].size();
216+
vector<int> heights(n);
217+
int ans = 0;
218+
for (auto& row : matrix)
219+
{
220+
for (int j = 0; j < n; ++j)
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{
222+
if (row[j] == '1') ++heights[j];
223+
else heights[j] = 0;
224+
}
225+
ans = max(ans, largestRectangleArea(heights));
226+
}
227+
return ans;
228+
}
229+
230+
int largestRectangleArea(vector<int>& heights) {
231+
int res = 0, n = heights.size();
232+
stack<int> stk;
233+
vector<int> left(n, -1);
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vector<int> right(n, n);
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for (int i = 0; i < n; ++i)
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{
237+
while (!stk.empty() && heights[stk.top()] >= heights[i])
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{
239+
right[stk.top()] = i;
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stk.pop();
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}
242+
if (!stk.empty()) left[i] = stk.top();
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stk.push(i);
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}
245+
for (int i = 0; i < n; ++i)
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res = max(res, heights[i] * (right[i] - left[i] - 1));
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return res;
248+
}
249+
};
250+
```
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252+
### **Go**
253+
254+
```go
255+
func maximalRectangle(matrix []string) int {
256+
if len(matrix) == 0 {
257+
return 0
258+
}
259+
n := len(matrix[0])
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heights := make([]int, n)
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ans := 0
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for _, row := range matrix {
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for j, v := range row {
264+
if v == '1' {
265+
heights[j]++
266+
} else {
267+
heights[j] = 0
268+
}
269+
}
270+
ans = max(ans, largestRectangleArea(heights))
271+
}
272+
return ans
273+
}
274+
275+
func largestRectangleArea(heights []int) int {
276+
res, n := 0, len(heights)
277+
var stk []int
278+
left, right := make([]int, n), make([]int, n)
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for i := range right {
280+
right[i] = n
281+
}
282+
for i, h := range heights {
283+
for len(stk) > 0 && heights[stk[len(stk)-1]] >= h {
284+
right[stk[len(stk)-1]] = i
285+
stk = stk[:len(stk)-1]
286+
}
287+
if len(stk) > 0 {
288+
left[i] = stk[len(stk)-1]
289+
} else {
290+
left[i] = -1
291+
}
292+
stk = append(stk, i)
293+
}
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for i, h := range heights {
295+
res = max(res, h*(right[i]-left[i]-1))
296+
}
297+
return res
298+
}
299+
300+
func max(a, b int) int {
301+
if a > b {
302+
return a
303+
}
304+
return b
305+
}
306+
```
307+
133308
### **...**
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135310
```

‎lcof2/剑指 Offer II 040. 矩阵中最大的矩形/Solution.go

+51
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@@ -0,0 +1,51 @@
1+
func maximalRectangle(matrix []string) int {
2+
if len(matrix) == 0 {
3+
return 0
4+
}
5+
n := len(matrix[0])
6+
heights := make([]int, n)
7+
ans := 0
8+
for _, row := range matrix {
9+
for j, v := range row {
10+
if v == '1' {
11+
heights[j]++
12+
} else {
13+
heights[j] = 0
14+
}
15+
}
16+
ans = max(ans, largestRectangleArea(heights))
17+
}
18+
return ans
19+
}
20+
21+
func largestRectangleArea(heights []int) int {
22+
res, n := 0, len(heights)
23+
var stk []int
24+
left, right := make([]int, n), make([]int, n)
25+
for i := range right {
26+
right[i] = n
27+
}
28+
for i, h := range heights {
29+
for len(stk) > 0 && heights[stk[len(stk)-1]] >= h {
30+
right[stk[len(stk)-1]] = i
31+
stk = stk[:len(stk)-1]
32+
}
33+
if len(stk) > 0 {
34+
left[i] = stk[len(stk)-1]
35+
} else {
36+
left[i] = -1
37+
}
38+
stk = append(stk, i)
39+
}
40+
for i, h := range heights {
41+
res = max(res, h*(right[i]-left[i]-1))
42+
}
43+
return res
44+
}
45+
46+
func max(a, b int) int {
47+
if a > b {
48+
return a
49+
}
50+
return b
51+
}

‎lcof2/剑指 Offer II 040. 矩阵中最大的矩形/Solution.java

+40
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@@ -0,0 +1,40 @@
1+
class Solution {
2+
public int maximalRectangle(String[] matrix) {
3+
if (matrix == null || matrix.length == 0) {
4+
return 0;
5+
}
6+
int n = matrix[0].length();
7+
int[] heights = new int[n];
8+
int ans = 0;
9+
for (var row : matrix) {
10+
for (int j = 0; j < n; ++j) {
11+
if (row.charAt(j) == '1') {
12+
heights[j] += 1;
13+
} else {
14+
heights[j] = 0;
15+
}
16+
}
17+
ans = Math.max(ans, largestRectangleArea(heights));
18+
}
19+
return ans;
20+
}
21+
22+
private int largestRectangleArea(int[] heights) {
23+
int res = 0, n = heights.length;
24+
Deque<Integer> stk = new ArrayDeque<>();
25+
int[] left = new int[n];
26+
int[] right = new int[n];
27+
Arrays.fill(right, n);
28+
for (int i = 0; i < n; ++i) {
29+
while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
30+
right[stk.pop()] = i;
31+
}
32+
left[i] = stk.isEmpty() ? -1 : stk.peek();
33+
stk.push(i);
34+
}
35+
for (int i = 0; i < n; ++i) {
36+
res = Math.max(res, heights[i] * (right[i] - left[i] - 1));
37+
}
38+
return res;
39+
}
40+
}

‎lcof2/剑指 Offer II 040. 矩阵中最大的矩形/Solution.py

+35
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@@ -0,0 +1,35 @@
1+
class Solution:
2+
def maximalRectangle(self, matrix: List[List[str]]) -> int:
3+
if not matrix:
4+
return 0
5+
heights = [0] * len(matrix[0])
6+
ans = 0
7+
for row in matrix:
8+
for j, v in enumerate(row):
9+
if v == "1":
10+
heights[j] += 1
11+
else:
12+
heights[j] = 0
13+
ans = max(ans, self.largestRectangleArea(heights))
14+
return ans
15+
16+
def largestRectangleArea(self, heights: List[int]) -> int:
17+
n = len(heights)
18+
stk = []
19+
left = [-1] * n
20+
right = [n] * n
21+
for i, h in enumerate(heights):
22+
while stk and heights[stk[-1]] >= h:
23+
stk.pop()
24+
if stk:
25+
left[i] = stk[-1]
26+
stk.append(i)
27+
stk = []
28+
for i in range(n - 1, -1, -1):
29+
h = heights[i]
30+
while stk and heights[stk[-1]] >= h:
31+
stk.pop()
32+
if stk:
33+
right[i] = stk[-1]
34+
stk.append(i)
35+
return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))

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