【leetcode】931. Minimum Falling Path Sum

本文详细解析了一道关于寻找矩阵中最小路径和的算法题目,通过动态规划的方法,定义了状态转移方程,并给出了具体的Python代码实现。示例中,对于给定的3x3矩阵,算法成功找到了最小路径和为12的路径。

题目如下:

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

解题思路:动态规划的入门级题目。状态转移方程: dp[i][j] = min(dp[i-1][j],dp[i-1][j-1],dp[i-1][j+1]) + A[i][j]。

代码如下:

class Solution(object):
    def minFallingPathSum(self, A):
        """
        :type A: List[List[int]]
        :rtype: int
        """
        for i in range(len(A)):
            for j in range(len(A[i])):
                if i - 1 < 0:
                    continue
                minv = A[i-1][j]
                if j - 1 >= 0:
                    minv = min(minv,A[i-1][j-1])
                if j + 1 < len(A[i]):
                    minv = min(minv,A[i-1][j+1])
                A[i][j] = minv + A[i][j]
        #print A
        return min(A[-1])

 

转载于:https://www.cnblogs.com/seyjs/p/9869071.html

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