[leetcode]Word Break @ Python

原题地址:https://oj.leetcode.com/problems/word-break/

题意:

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

解题思路:这道题考察的显然不是dfs,为什么?因为这道题不需要给出如何分割的答案,只需要判断是否可以分割为字典中的单词即可。我们考虑使用动态规划,这个思路看代码的话不难,用python写起来也比较清晰。

代码:

[leetcode]Word Break @ Python
class Solution:
    # @param s, a string
    # @param dict, a set of string
    # @return a boolean
    # @good coding!
    def wordBreak(self, s, dict):
        dp = [False for i in range(len(s)+1)]
        dp[0] = True
        for i in range(1, len(s)+1):
            for k in range(i):
                if dp[k] and s[k:i] in dict:
                    dp[i] = True
        return dp[len(s)]
[leetcode]Word Break @ Python

 

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[leetcode]Word Break @ Python

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