Leetcode 139. Word Break

最新推荐文章于 2025-09-06 22:33:46 发布

weixin_30734435

最新推荐文章于 2025-09-06 22:33:46 发布

阅读量57

点赞数

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/A-Little-Nut/p/10055778.html

本文探讨了如何使用动态规划解决字符串分割问题，具体为判断一个非空字符串是否可以被分割成一个或多个字典中出现的单词序列。通过一个具体的例子，详细解释了动态规划算法的实现过程。

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

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

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

分析

动态规划

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
         vector<int> dp(s.size()+1,0);
         dp[0]=1;
         for(int i=0;i<s.size();i++)
             for(int j=i;j>=0;j--)
                 for(int k=0;k<wordDict.size();k++)
                     if(wordDict[k]==s.substr(j,i-j+1)&&dp[j]==1)
                        dp[i+1]=1;
        if(dp[s.size()]==1)
            return true;
        else
            return false;
    }
};

转载于:https://www.cnblogs.com/A-Little-Nut/p/10055778.html