2019.7.1 感谢张老大付费的马拉维的小孩的视频,张老大牛批!!!
题目地址:Maximal Rectangle
题目简介:
给定一个仅包含 0 和 1 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。
Input:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
Output: 6
题目解析:
按照最大正方形的方法解决问题不好解决,需要动态的更新宽和高。根据discuss区,看到使用Largest Rectangle in Histogram方法。即将matrix的没一层看作底,计算最大直方图。(真不知道如果不是刚做84怎么想起来的)
C++:
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
if (matrix.size() == 0 || matrix[0].size() == 0)
return 0;
int _size = matrix[0].size();
vector<int> height(_size + 1, 0);
int ans = 0;
for (int i = 0; i < matrix.size(); i++)
{
for (int j = 0; j < _size; j++)
height[j] = matrix[i][j] == '1' ? height[j] + 1 : 0;
stack<int> st;
st.push(-1);
for (int k = 0; k < _size + 1; k++)
{
while((st.top() != -1 && height[k] < height[st.top()]) || (height[k] < height[_size]))
{
int h = height[st.top()];
st.pop();
int w = k - 1 - st.top();
ans = max(ans, h * w);
}
st.push(k);
}
}
return ans;
}
};
Python:
class Solution:
def maximalRectangle(self, matrix):
if not matrix or not matrix[0]:
return 0
n = len(matrix[0])
height = [0] * (n + 1)
ans = 0
for row in matrix:
for i in range(n):
height[i] = height[i] + 1 if row[i] == '1' else 0
stack = [-1]
for i in range(n + 1):
while height[i] < height[stack[-1]]:
h = height[stack.pop()]
w = i - 1 - stack[-1]
ans = max(ans, h * w)
stack.append(i)
return ans