标签：value current led href sid statement ini ++ lex
Problem statement:
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square containing only 1‘s and return its area.
For example, given the following matrix:
1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0
Return 4.
Solution:
It looks like 85. Maximal Rectangle. But, there is big difference. This is DP problem, however, maximul rectangle needs tricky.
The key point is what we want from DP? The answer is max side length of square.
DP maintains a 2D array, dp[i][j] means the max side of length of square by current position.
Initialization:
dp[0][j] = matrix[0][j] - ‘0‘; dp[i][0] = matrix[i][0] - ‘0‘;
DP formula. and update the max side length for each value.
if(matrix[i][j] == ‘1‘){ dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; }
Return max_size * max_size;
The time complexity is O(n * n)
class Solution { public: int maximalSquare(vector<vector<char>>& matrix) { if(matrix.empty()){ return 0; } int row = matrix.size(); int col = matrix[0].size(); vector<vector<int>> square_size(row, vector<int>(col, 0)); int max_size = 0; for(int i = 0; i < row; i++){ square_size[i][0] = matrix[i][0] - ‘0‘; max_size = max(max_size, square_size[i][0]); } for(int j = 0; j < col; j++){ square_size[0][j] = matrix[0][j] - ‘0‘; max_size = max(max_size, square_size[0][j]); } for(int i = 1; i < row; i++){ for(int j = 1; j < col; j++){ if(matrix[i][j] == ‘1‘){ square_size[i][j] = min(square_size[i - 1][j - 1], min(square_size[i - 1][j], square_size[i][j - 1])) + 1; } max_size = max(max_size, square_size[i][j]); } } return max_size * max_size; } };
标签：value current led href sid statement ini ++ lex
原文地址：http://www.cnblogs.com/wdw828/p/6851507.html
