Google编程大赛入围赛750分真题 第五组

本文介绍了Google编程大赛入围赛750分真题第五组。题目要求在给定的字母网格中寻找指定单词,规定了路径移动规则和返回结果条件。还给出了不同网格和单词组合的示例,展示了如何计算找到单词的路径数量。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Google编程大赛入围赛750分真题 第五组


  Problem Statement
  牋牋

  You are given a String[] grid representing a rectangular grid of letters. You
  
are also given a String find, a word you are to find within the grid. The
  
starting point may be anywhere in the grid. The path may move up, down, left,
  
right, or diagonally from one letter to the next, and may use letters in the
  
grid more than once, but you may not stay on the same cell twice in a row (see
  
example 6 for clarification). You are to return an int indicating the number of
  
ways find can be found within the grid. If the result is more than
  
1,000,000,000, return -1. Definition
  牋牋

  Class:
  
WordPath
  
Method:
  
countPaths
  
Parameters:
  
String[], String
  
Returns:
  
int
  
Method signature:
  
int countPaths(String[] grid, String find)
  
(be sure your method is public)
  牋牋


  Constraints
  
-
  
grid will contain between 1 and 50 elements, inclusive.
  
-
  
Each element of grid will contain between 1 and 50 uppercase ('A'-'Z') letters, inclusive.
  
-
  
Each element of grid will contain the same number of characters.
  
-
  
find will contain between 1 and 50 uppercase ('A'-'Z') letters, inclusive.
  
Examples
  
0)

  牋牋

  {"ABC",
  
"FED",
  
"GHI"}
  
"ABCDEFGHI"
  
Returns: 1
  
There is only one way to trace this path. Each letter is used exactly once.
  
1)

  牋牋

  {"ABC",
  
"FED",
  
"GAI"}
  
"ABCDEA"
  
Returns: 2
  
Once we get to the 'E', we can choose one of two directions for the final 'A'.
  
2)

  牋牋

  {"ABC",
  
"DEF",
  
"GHI"}
  
"ABCD"
  
Returns: 0
  
We can trace a path for "ABC", but there's no way to complete a path to the letter 'D'.
  
3)

  牋牋

  {"AA",
  
"AA"}
  
"AAAA"
  
Returns: 108
  
We can start from any of the four locations. From each location, we can then
  
move in any of the three possible directions for our second letter, and again
  
for the third and fourth letter. 4 * 3 * 3 * 3 = 108. 4)

  牋牋

  {"ABABA",
  
"BABAB",
  
"ABABA",
  
"BABAB",
  
"ABABA"}
  
"ABABABBA"
  
Returns: 56448
  
There are a lot of ways to trace this path.
  
5)

  牋牋

  {"AAAAA",
  
"AAAAA",
  
"AAAAA",
  
"AAAAA",
  
"AAAAA"}
  
"AAAAAAAAAAA"
  
Returns: -1
  
There are well over 1,000,000,000 paths that can be traced.
  
6)

  牋牋

  {"AB",
  
"CD"}
  
"AA"
  
Returns: 0
  
Since we can't stay on the same cell, we can't trace the path at all.


  
This problem statement is the exclusive and proprietary property of TopCoder,
  
Inc. Any unauthorized use or reproduction of this information without the prior
  
written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder,
  Inc. All rights reserved.

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值