931. Minimum Falling Path Sum

最新推荐文章于 2022-08-26 00:33:12 发布

SnailRunFast

最新推荐文章于 2022-08-26 00:33:12 发布

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本文探讨了在给定的正方形整数数组中寻找最小下降路径和的问题，详细解释了动态规划方法的应用，通过实例展示了如何从第一行开始，每次选择下一列中相邻元素的最小和路径。

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931. Minimum Falling Path Sum

Medium

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

Solution: DP array[i][j] += Min(array[i - 1][j], array[i - 1][Max(0, j - 1)], array[i - 1][Min(columns - 1, j + 1)]); i > 1

then get the minimum element of the last row.

class Solution {
    public int minFallingPathSum(int[][] A) {
        int row = A.length;
        int column = A[0].length;
        
        for (int i = 1; i < row; i++) {
            for (int j = 0; j < column; j++) {
                A[i][j] += Math.min(A[i - 1][j], 
                                Math.min(A[i - 1][Math.max(0, j - 1)], 
                                            A[i - 1][Math.min(column - 1, j + 1)]));
            }
        }
        int minVal = A[row - 1][0];
        
        for (int i = 0; i < column; i++) {
            minVal = Math.min(minVal, A[row - 1][i]);
        }
        
        return minVal;
    }
}