LeetCode 931. Minimum Falling Path Sum

[Woodyiiiiiii]

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

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很简单。二维dp数组如下：

class Solution {
    public int minFallingPathSum(int[][] A) {
        int row = A.length;
        int col = A[0].length;
        int[][] dp = new int[row][col];
        int res = Integer.MAX_VALUE, i = 0, j = 0;
        for (i = 0; i < col; ++i) {
            dp[0][i] = A[0][i];
        }

        for (i = 1; i < row; ++i) {
            for (j = 0; j < col; ++j) {
                if (j == 0) {
                    dp[i][j] = A[i][j] + Math.min(dp[i - 1][j], dp[i - 1][j + 1]);
                }
                else if (j == col - 1) {
                    dp[i][j] = A[i][j] + Math.min(dp[i - 1][j], dp[i - 1][j - 1]);
                }
                else  {
                    // Math.min(a, b) in java.lang
                    dp[i][j] = A[i][j] + Math.min(dp[i - 1][j],
                            Math.min(dp[i - 1][j - 1], dp[i - 1][j + 1]));
                }
            }
        }


        for (i = 0; i < col; ++i) {
            res = (dp[row - 1][i] < res) ? dp[row - 1][i] : res;
        }
        return res;
    }
}

当然也可以优化成一维数组(需要声明一个变量保存初值)，这里不再赘述。

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参考资料:
LeetCode原题