221 medium
221. Maximal Square
Medium
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square containing only 1‘s and return its area.
Example:
Input: 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0 Output: 4
To appy DP, we define the state as the maximal size (square = size * size) of the square that can be formed till point (i, j), denoted as dp[i][j].
For the topmost row (i = 0) and the leftmost column (j = 0), we have dp[i][j] = matrix[i][j] - ‘0‘, meaning that it can at most form a square of size 1 when the matrix has a ‘1‘in that cell.
When i > 0 and j > 0, if matrix[i][j] = ‘0‘, then dp[i][j] = 0 since no square will be able to contain the ‘0‘ at that cell. If matrix[i][j] = ‘1‘, we will have dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1, which means that the square will be limited by its left, upper and upper-left neighbors.
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.empty()) {
return 0;
}
int m = matrix.size(), n = matrix[0].size(), sz = 0;
vector<vector<int>> dp(m, vector<int>(n, 0));
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!i || !j || matrix[i][j] == ‘0‘) {
dp[i][j] = matrix[i][j] - ‘0‘;
} else {
dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
}
sz = max(dp[i][j], sz);
}
}
return sz * sz;
}
};
In the above code, it uses O(mn) space. Actually each time when we update dp[i][j], we only need dp[i-1][j-1], dp[i-1][j] (the previous row) and dp[i][j-1] (the current row). So we may just keep two rows.
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.empty()) {
return 0;
}
int m = matrix.size(), n = matrix[0].size(), sz = 0;
vector<int> pre(n, 0), cur(n, 0);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!i || !j || matrix[i][j] == ‘0‘) {
cur[j] = matrix[i][j] - ‘0‘;
} else {
cur[j] = min(pre[j - 1], min(pre[j], cur[j - 1])) + 1;
}
sz = max(cur[j], sz);
}
fill(pre.begin(), pre.end(), 0);
swap(pre, cur);
}
return sz * sz;
}
};
Furthermore, we may only use just one vector (thanks to @stellari for sharing the idea).
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.empty()) {
return 0;
}
int m = matrix.size(), n = matrix[0].size(), sz = 0, pre;
vector<int> cur(n, 0);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int temp = cur[j];
if (!i || !j || matrix[i][j] == ‘0‘) {
cur[j] = matrix[i][j] - ‘0‘;
} else {
cur[j] = min(pre, min(cur[j], cur[j - 1])) + 1;
}
sz = max(cur[j], sz);
pre = temp;
}
}
return sz * sz;
}
};
原文地址：https://www.cnblogs.com/shona/p/11311161.html