Leetcode Word Break-优快云博客

本文链接：https://blog.youkuaiyun.com/rosepicker/article/details/75040103

本文介绍了一个经典的字符串分割匹配问题，并采用动态规划的方法进行求解。给出了一段C++实现代码，利用动态规划判断一个字符串是否可以被字典中的单词完全匹配。

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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".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

这题目一看就是动态规划大法，该动态规划方程与一般的有点小区别，动态规划方程如下：

初始化：dp[s.size()+1], dp[0] = true

j = 0,1,2,...,i-1,if(dp[j] && s.substr(j,i-j) in wordDict) dp[i] = true;

代码如下：

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        bool* dp = new bool[s.size()+1];
        memset(dp,false,sizeof(bool)*(s.size()+1));
        dp[0] = true;
        
        for(int i=1;i<=s.size();i++)
            for(int j=i-1;j>=0;j--)
            {
                string tmp = s.substr(j,i-j);
                if(dp[j] && find(wordDict.begin(),wordDict.end(),tmp) != wordDict.end())
                {
                    dp[i] = true;
                    break;
                }
            }
        return dp[s.size()];
    }
};