27
07/2015
[LeetCode] Maximal Square
Maximal Square
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 0Return 4.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
解题思路：
最值问题考虑动态规划算法。设d[i][j]表示以matrix[i][j]为右下角的最大方阵大小，则有：
d[i][j] = 0; matrix[i][j]='0'时
d[i][j] = 1; matrix[i][j]='1'且(i==0||j==0)时
d[i][j] = pow(min(min(sqrt(d[i-1][j-1]), sqrt(d[i-1][j])), sqrt(d[i][j-1])) + 1, 2); 其他情况。
然后用一个变量maxSquare记录全局最大的方阵即可。
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
int n = matrix.size();
if(n==0){
return 0;
}
int m = matrix[0].size();
vector<vector<int>> d(n, vector<int>(m, 0)); //以matrix[i][j]为右下角的最大方阵大小
int maxSquare=0;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(matrix[i][j]=='0'){
d[i][j] = 0;
}else{
if(i>0&&j>0){
d[i][j] = (int)pow(min((int)min((int)sqrt(d[i-1][j-1]), (int)sqrt(d[i-1][j])), (int)sqrt(d[i][j-1])) + 1, 2);
}else{
d[i][j] = 1;
}
maxSquare = max(maxSquare, d[i][j]);
}
}
}
return maxSquare;
}
};转载请注明：康瑞部落 » [LeetCode] Maximal Square
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