原题地址:https://leetcode-cn.com/problems/maximal-rectangle/description/
题目描述:
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。
示例:
输入:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
输出: 6
解题方案:
参考链接:https://blog.csdn.net/LaputaFallen/article/details/79869806 该链接包含正方形矩阵和矩形矩阵的解题思路,本题是矩形矩阵更难一些。运用到了动态规划,现在就还没完全搞明白这道题的解题方式。。。留着以后再学习。
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
if(matrix.empty()) return 0;
int m = matrix.size();
int n = matrix[0].size();
int left[n], right[n], height[n];
fill_n(left, n, 0);
fill_n(right, n, n);
fill_n(height, n, 0);
int maxA = 0;
for(int i = 0; i < m; i ++) {
int cur_left = 0, cur_right = n;
for(int j = 0; j < n; j ++) { // compute height (can do this from either side)
if(matrix[i][j] == '1') height[j] ++;
else height[j] = 0;
}
for(int j = 0; j < n; j ++) { // compute left (from left to right)
if(matrix[i][j] == '1') left[j] = max(left[j], cur_left);
else {
left[j] = 0;
cur_left = j + 1;
}
}
for(int j = n - 1; j >= 0; j --) { // compute right (from right to left)
if(matrix[i][j] == '1') right[j] = min(right[j], cur_right);
else {
right[j]=n;
cur_right=j;
}
}
// compute the area of rectangle (can do this from either side)
for(int j = 0; j < n; j ++)
maxA = max(maxA, (right[j] - left[j]) * height[j]);
}
return maxA;
}
};