LeetCode 931. 下降路径最小和（动态规划）

示例：
输入：[[1,2,3],[4,5,6],[7,8,9]]
输出：12
解释：
可能的下降路径有：
[1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
[2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
[3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]
和最小的下降路径是 [1,4,7]，所以答案是 12。

提示：
1 <= A.length == A[0].length <= 100
-100 <= A[i][j] <= 100

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        const int N = A.size();
        vector<vector<int>> dp(N,vector<int>(N+2,INT_MAX));
        int i, j, minSum = INT_MAX;
        for(i = 0; i < N; i++) 
        	dp[0][i+1] = A[0][i];//初始化第一行
        for(i = 1; i < N; ++i)
        {
        	for(j = 1; j < N+1; ++j)
        	{
        		dp[i][j] = A[i][j-1]+min(min(dp[i-1][j-1],dp[i-1][j]),dp[i-1][j+1]);
        	}
        }
        for(i = 1; i < N+1; ++i)
        	minSum = min(minSum,dp[N-1][i]);
        return minSum;
    }
};

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        const int N = A.size();
        vector<int> dp(N+2,INT_MAX);
        vector<int> temp(N+2,INT_MAX);
        int i, j, minSum = INT_MAX;
        for(i = 0; i < N; i++) 
            temp[i+1] = A[0][i];//初始化第一行
        for(i = 1; i < N; ++i)
        {
            for(j = 1; j < N+1; ++j)
            {   //注意加了两列后下标的错位
                dp[j] = A[i][j-1]+min(min(temp[j-1],temp[j]),temp[j+1]);
            }
            swap(dp,temp);
        }
        return *min_element(temp.begin(),temp.end());
    }
};