重点：要仅在matrix[i][j] == 1时，更新状态方程，跳过等于0.
同时: 可以用[row+1][col+1]的方法来实现，并初始化[0][0]为0， 使得代码简洁。
int maximalSquare(vector<vector<char>>& matrix) {
if(matrix.empty() || matrix[0].empty()) return 0;
int row = matrix.size(), col = matrix[0].size();
vector<vector<int>> dp(row+1, vector<int>(col+1, 0));
int max_len = 0;
for(int i=1; i<=row; i++){
for(int j=1; j<=col; j++){
if(matrix[i-1][j-1] == '1'){
dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1;
max_len = max(max_len, dp[i][j]);
}
}
}
return max_len * max_len;
}
sliding array to further reduce space complexity:
滚动数组的优化：注意，由于滚动数组并不能够做到对all row and all column的初始化了，所以matrix[i][j] == 0时不能跳过，要将dp[i][j]设为0；
class Solution {
public:
/**
* @param matrix: a matrix of 0 and 1
* @return: an integer
*/
int maxSquare(vector<vector<int> > &matrix) {
// write your code here
if(matrix.empty() || matrix[0].empty()) return 0;
int row = matrix.size(), col = matrix[0].size();
vector<vector<int>> dp(2, vector<int>(col+1, 0));
int max_len = 0;
for(int i=1; i<=row; i++){
for(int j=1; j<=col; j++){
if(matrix[i-1][j-1] == 1){
dp[i%2][j] = min(dp[(i-1)%2][j-1], min(dp[(i-1)%2][j], dp[i%2][j-1])) + 1;
if(dp[i%2][j] > max_len) max_len = dp[i%2][j];
}else{
dp[i%2][j] = 0;
}
}
}
return max_len * max_len;
}
};
