leetcode #139 in cpp_139cpp-优快云博客

本文探讨了如何判断一个字符串是否能被分割成一个或多个字典中出现的单词序列。提供了两种解决方案：递归方法（带有剪枝）和动态规划方法，并详细展示了每种方法的实现过程。

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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".

Code 1:(recursive with pruning)

class Solution {
public:
    bool wordBreak(string s, unordered_set<string>& wordDict) {
        if(s.empty()) return wordDict.find("") == wordDict.end()?false:true;
      
        int end = 0; 
        while( start < s.length() && end < s.length()){
            if(wordDict.find(s.substr(start,end - start + 1)) == wordDict.end()){
                end++; 
            }else{
                start = end + 1;
                end = start; 
            }
        }
        return start == end;
    }
    
    bool recursive_with_prune(int start, int end, string &s, unordered_set<string>& wordDict, vector<vector<int>> &res){
        
        if(start >= s.length()){
            return true;
        }
        if(end < s.length() && res[start][end] != -1){
            return res[start][end] == 1?true:false;
        }
        for(int i = end; i < s.length(); i ++){
            if(wordDict.find(s.substr(start, i- start + 1)) != wordDict.end() && recursive_with_prune(i+1, i+1, s, wordDict,res)){
                return true;
            }
        }
        res[start][end] = false; 
        return false; 
    }
};

Code 2: (DP)

class Solution {
public:
    bool wordBreak(string s, unordered_set<string>& wordDict) {
        if(s.empty()) return wordDict.find("") == wordDict.end()?false:true;
        return dp(s,wordDict); 
    }
    bool dp(string &s, unordered_set<string>& wordDict){
        int n = s.length(); 
        vector<bool> dp(n,false); //dp[i] means s[0....i] can formed by the dict 
        for(int i = 0 ; i < n; i ++){ 
            if(wordDict.find(s.substr(0,i+1))!=wordDict.end()) dp[i] = true; 
            else{
                for(int j = 0; j < i; j ++){//if s[0...j] can be formed by dict and s[j+1,..,i] is in the dict, then dp[i] = true
                    if(dp[j] && wordDict.find(s.substr(j+1,i-j))!=wordDict.end())
                        dp[i] = true; 
                }
            }
        }
        return dp[n-1]; 
        
    }
};