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JobHunting版 - LeetCode: Word Ladder
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相关话题的讨论汇总
话题: word话题: given话题: length话题: words
进入JobHunting版参与讨论
1 (共1页)
j********r
发帖数: 25
1
题目:
Given two words (start and end), and a dictionary, find the length of
shortest transformation sequence from start to end, such that:
Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
为什么如下distance 是 2 而不是 1?
Submission Result: Wrong Answer
Input: "a", "c", ["a","b","c"]
Output: 1
Expected:2
l*n
发帖数: 529
2
到底是1还是2无所谓,定义上一致就行。题目要求的其实是变换路径的节点数,从a->c
就是两个节点,所以是2.给的例子结果是5也是算的路径节点数而不是jump数。

【在 j********r 的大作中提到】
: 题目:
: Given two words (start and end), and a dictionary, find the length of
: shortest transformation sequence from start to end, such that:
: Only one letter can be changed at a time
: Each intermediate word must exist in the dictionary
: For example,
: Given:
: start = "hit"
: end = "cog"
: dict = ["hot","dot","dog","lot","log"]

K***s
发帖数: 621
3
start就算1,你仔细读题

【在 j********r 的大作中提到】
: 题目:
: Given two words (start and end), and a dictionary, find the length of
: shortest transformation sequence from start to end, such that:
: Only one letter can be changed at a time
: Each intermediate word must exist in the dictionary
: For example,
: Given:
: start = "hit"
: end = "cog"
: dict = ["hot","dot","dog","lot","log"]

j********r
发帖数: 25
4
确实是这样, 谢谢。

【在 K***s 的大作中提到】
: start就算1,你仔细读题
1 (共1页)
进入JobHunting版参与讨论
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相关话题的讨论汇总
话题: word话题: given话题: length话题: words