题目描述
给定一个由 0 和 1 组成的矩阵 matrix ，找出只包含 1 的最大矩形，并返回其面积。
注意：此题 matrix 输入格式为一维 01 字符串数组。
示例 1：
输入：matrix = ["10100","10111","11111","10010"]
输出：6
解释：最大矩形如上图所示。
示例 2：
输入：matrix = []
输出：0
示例 3：
输入：matrix = ["0"]
输出：0
示例 4：
输入：matrix = ["1"]
输出：1
示例 5：
输入：matrix = ["00"]
输出：0
提示：
rows == matrix.lengthcols == matrix[0].length0 <= row, cols <= 200matrix[i][j] 为 '0' 或 '1'
注意：本题与主站 85 题相同（输入参数格式不同）： https://leetcode.cn/problems/maximal-rectangle/
解法
方法一：单调栈
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。
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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35 | class Solution:
def maximalRectangle(self, matrix: List[List[str]]) -> int:
if not matrix:
return 0
heights = [0] * len(matrix[0])
ans = 0
for row in matrix:
for j, v in enumerate(row):
if v == "1":
heights[j] += 1
else:
heights[j] = 0
ans = max(ans, self.largestRectangleArea(heights))
return ans
def largestRectangleArea(self, heights: List[int]) -> int:
n = len(heights)
stk = []
left = [-1] * n
right = [n] * n
for i, h in enumerate(heights):
while stk and heights[stk[-1]] >= h:
stk.pop()
if stk:
left[i] = stk[-1]
stk.append(i)
stk = []
for i in range(n - 1, -1, -1):
h = heights[i]
while stk and heights[stk[-1]] >= h:
stk.pop()
if stk:
right[i] = stk[-1]
stk.append(i)
return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))
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40 | class Solution {
public int maximalRectangle(String[] matrix) {
if (matrix == null || matrix.length == 0) {
return 0;
}
int n = matrix[0].length();
int[] heights = new int[n];
int ans = 0;
for (var row : matrix) {
for (int j = 0; j < n; ++j) {
if (row.charAt(j) == '1') {
heights[j] += 1;
} else {
heights[j] = 0;
}
}
ans = Math.max(ans, largestRectangleArea(heights));
}
return ans;
}
private int largestRectangleArea(int[] heights) {
int res = 0, n = heights.length;
Deque<Integer> stk = new ArrayDeque<>();
int[] left = new int[n];
int[] right = new int[n];
Arrays.fill(right, n);
for (int i = 0; i < n; ++i) {
while (!stk.isEmpty() && heights[stk.peek()] >= heights[i]) {
right[stk.pop()] = i;
}
left[i] = stk.isEmpty() ? -1 : stk.peek();
stk.push(i);
}
for (int i = 0; i < n; ++i) {
res = Math.max(res, heights[i] * (right[i] - left[i] - 1));
}
return res;
}
}
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31 | class Solution {
public:
int h[210];
int l[210], r[210];
int maximalRectangle(vector<string>& matrix) {
int n = matrix.size();
if (n == 0) return 0;
int m = matrix[0].size();
int ans = 0;
stack<int> st;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
h[j] = (matrix[i][j] == '1' ? h[j] + 1 : 0);
while (st.size() && h[j] <= h[st.top()]) {
ans = max(ans, (j - l[st.top()] - 1) * h[st.top()]);
st.pop();
}
if (st.size())
l[j] = st.top();
else
l[j] = -1;
st.push(j);
}
while (st.size()) {
ans = max(ans, (m - 1 - l[st.top()]) * h[st.top()]);
st.pop();
}
}
return ans;
}
};
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44 | func maximalRectangle(matrix []string) int {
if len(matrix) == 0 {
return 0
}
n := len(matrix[0])
heights := make([]int, n)
ans := 0
for _, row := range matrix {
for j, v := range row {
if v == '1' {
heights[j]++
} else {
heights[j] = 0
}
}
ans = max(ans, largestRectangleArea(heights))
}
return ans
}
func largestRectangleArea(heights []int) int {
res, n := 0, len(heights)
var stk []int
left, right := make([]int, n), make([]int, n)
for i := range right {
right[i] = n
}
for i, h := range heights {
for len(stk) > 0 && heights[stk[len(stk)-1]] >= h {
right[stk[len(stk)-1]] = i
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
left[i] = stk[len(stk)-1]
} else {
left[i] = -1
}
stk = append(stk, i)
}
for i, h := range heights {
res = max(res, h*(right[i]-left[i]-1))
}
return res
}
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方法二