Epic - Snake Sequence

最新推荐文章于 2025-08-21 10:38:03 发布

weixin_30919919

最新推荐文章于 2025-08-21 10:38:03 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/lilixu/p/4574745.html

本文介绍了一种基于动态规划算法解决寻找网格中蛇形数列的问题。蛇形数列是指网格中的相邻数字序列，每个数字与其右侧或下方的数字之差的绝对值为1。文章提供了详细的算法实现步骤，并通过一个具体的示例来展示如何找到最长的蛇形数列及其长度。

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You are given a grid of numbers. A snakes equence is made up of adjacent numbers such that for each number, the number on the right or the number below it is +1 or -1 its value. For example,

1 3 2 6 8

-9 7 1 -1 2

1 5 0 1 9

In this grid, (3, 2, 1, 0, 1) is a snake sequence. Given a grid, find the longest snake sequences and their lengths (so there can be multiple snake sequences with the maximum length).

简单的动态规划，使用一个相同大小的数组记录到snake sequence该点的长度（默认为1），并维护最大值。

def snake(matrix)
  return 0 if matrix.empty?
  m , n = matrix.length, matrix[0].length
  dp = Array.new(m){Array.new(n,1)}
  ans = -1
  m.times do |i|
    n.times do |j|
      dp[i][j] = dp[i-1][j] + 1 if i > 0 and (matrix[i-1][j] - matrix[i][j]).abs == 1
      dp[i][j] = dp[i][j-1] + 1 if j > 0 and (matrix[i][j-1] - matrix[i][j]).abs == 1
      ans = [ans,dp[i][j]].max
    end
  end
  ans
end