### 题目

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

### 典型的动态规划问题

1,  2,  3
4,  5,  6
7,  8,  9


1,  2,  3
\ | /
5


f(4),   f(5),   f(6)
\    |    /
8


1,  2,  3
5,  6,  8
12, 12, 15


#### 代码

class Solution {
public int minFallingPathSum(int[][] A) {
int height = A.length, width = A[0].length;
int[] dpRow = Arrays.copyOf(A[0], width);
for (int i = 1; i < height; i++) {
int[] nextRow = new int[width];
for (int j = 0; j < width; j++) {
nextRow[j] = dpRow[j] + A[i][j];
if (j > 0) nextRow[j] = Math.min(nextRow[j], dpRow[j - 1] + A[i][j]);
if (j < width - 1) nextRow[j] = Math.min(nextRow[j], dpRow[j + 1] + A[i][j]);
}
dpRow = nextRow;
}
int res = 10001;
for (int n : dpRow) res = Math.min(res, n);
return res;
}
}