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Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square containing all 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.
Runtime: 12ms.
1 class Solution { 2 public: 3 int maximalSquare(vector<vector<char>>& matrix) { 4 if(matrix.empty() || matrix[0].empty()) return 0; 5 6 int row = matrix.size(); 7 int col = matrix[0].size(); 8 9 vector<vector<int> > dp(row, vector<int>(col, 0)); 10 int result = 0; 11 for(int i = 0; i < row; i++) 12 if(matrix[i][0] == ‘1‘){ 13 dp[i][0] = 1; 14 result = 1; 15 } 16 17 for(int j = 0; j < col; j++) 18 if(matrix[0][j] == ‘1‘){ 19 dp[0][j] = 1; 20 result = 1; 21 } 22 23 for(int i = 1; i < row; i++){ 24 for(int j = 1; j < col; j++){ 25 if(matrix[i][j] == ‘1‘){ 26 dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; 27 result = max(result, dp[i][j]); 28 } 29 } 30 } 31 return result * result; 32 } 33 };class Solution { 34 public: 35 int maximalSquare(vector<vector<char>>& matrix) { 36 if(matrix.empty() || matrix[0].empty()) return 0; 37 38 int row = matrix.size(); 39 int col = matrix[0].size(); 40 41 vector<vector<int> > dp(row, vector<int>(col, 0)); 42 int result = 0; 43 for(int i = 0; i < row; i++) 44 if(matrix[i][0] == ‘1‘){ 45 dp[i][0] = 1; 46 result = 1; 47 } 48 49 for(int j = 0; j < col; j++) 50 if(matrix[0][j] == ‘1‘){ 51 dp[0][j] = 1; 52 result = 1; 53 } 54 55 for(int i = 1; i < row; i++){ 56 for(int j = 1; j < col; j++){ 57 if(matrix[i][j] == ‘1‘){ 58 dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; 59 result = max(result, dp[i][j]); 60 } 61 } 62 } 63 return result * result; 64 } 65 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4845396.html
