967. Minimum Falling Path Sum

难度: Medium

题目描述

Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix.

A falling path starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1).

 

Example 1:

Input: matrix = [[2,1,3],[6,5,4],[7,8,9]]
Output: 13
Explanation: There are two falling paths with a minimum sum as shown.

Example 2:

Input: matrix = [[-19,57],[-40,-5]]
Output: -59
Explanation: The falling path with a minimum sum is shown.

 

Constraints:

• n == matrix.length == matrix[i].length
• 1 <= n <= 100
• -100 <= matrix[i][j] <= 100

解题思路

代码实现

解决方案

java

class Solution {
    public int minFallingPathSum(int[][] matrix) {
        if (matrix.length == 1) {
            return matrix[0][0];
        }
        int n = matrix.length;
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0) {
                    dp[i][j] = matrix[i][j];
                    continue;
                }
                if (j == 0 && j < n - 1) {
                    dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j + 1]) + matrix[i][j];
                    continue;
                }
                if (j == n - 1 && j > 0) {
                    dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j - 1]) + matrix[i][j];
                    continue;
                }
                dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i - 1][j - 1]), dp[i - 1][j + 1]) + matrix[i][j];

            }
        }
        Arrays.sort(dp[n - 1]);
        return dp[n - 1][0];

    }
}