931. Minimum Falling Path Sum

Given a square array of integers A, we want the minimum sum of a falling path through A.

A falling path starts at any element in the first row, and chooses one element from each row.  The next row's choice must be in a column that is different from the previous row's column by at most one.

Example 1:

Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: 12
Explanation: 
The possible falling paths are:

• [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]

The falling path with the smallest sum is [1,4,7], so the answer is 12.

Note:

1. 1 <= A.length == A[0].length <= 100
2. -100 <= A[i][j] <= 100

Approath #1: Bottom to Top. [C++]

class Solution {
    public int minFallingPathSum(int[][] A) {
        int l = A.length;
        int[][] dp = new int[l+1][l+1];
        for (int i = 0; i < l; ++i)
            for (int j = 0; j < l; ++j)
                dp[i][j] = A[i][j];
        for (int i = l-2; i >= 0; --i) {
            for (int j = 0; j < l; ++j) {
                int left = j > 0 ? dp[i+1][j-1] : Integer.MAX_VALUE;
                int right = j < l-1 ? dp[i+1][j+1] : Integer.MAX_VALUE;
                int down = dp[i+1][j];
                dp[i][j] += Math.min(left, Math.min(down, right));
                // System.out.print("dp[" + i + "][" + j + "]= " + dp[i][j] + " ");
            }
            // System.out.println();
        }
        
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < l; ++i)
            ans = Math.min(ans, dp[0][i]);
        
        return ans;
    }
}

　　

Analysis:

Solving this problem using the thought of 'bottom to top', we calculate the minimum sum using dp[i][j] = dp[i][j] + min(left, right, down).

Finally,  we can find the answer at the first row.