【網格 dp】B004_LC_下降路徑最小和 I（邊界判斷）

一、Problem

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.

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.

Constraints:

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

二、Solution

方法一：dp

• 定義狀態：
  • $f[i][j]$ 第 $i$ 行選第 $j$ 個數時的最小路徑和
• 思考初始化：
  • $f[i][j] = A[i][j]$
• 思考狀態轉移方程：
  • $f[i][j] = A[i][j] + min(f[i-1][j-1], f[i-1][j], f[i-1][j+1])$
• 思考輸出：$min(f[n-1][0...n])$

class Solution {
    public int minFallingPathSum(int[][] A) {
        int n = A.length, m = A[0].length, f[][] = A, INF = 0x3f3f3f3f;

        for (int i = 1; i < n; i++)
        for (int j = 0; j < m; j++) {
            if (j == 0)
                f[i][j] += Math.min(f[i-1][j], f[i-1][j+1]);
            else if (j == m-1)
                f[i][j] += Math.min(f[i-1][j], f[i-1][j-1]);
            else
                f[i][j] += Math.min(f[i-1][j-1], Math.min(f[i-1][j], f[i-1][j+1]));
        }
        int min = INF;
        for (int x : f[n-1]) if (min > x)
            min = x;
        return min;
    }
}

複雜度分析

• 時間複雜度：$O(n^2)$，
• 空間複雜度：$O(n^2)$，