【LeetCode】139. Word Break

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.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

链接：https://leetcode.com/problems/word-break

题解：跟上一篇一样，难度更简单，直接遍历s求出dp数组就行，dp记录每个位置之前的字符串是否可分，以后bug free要注意调用函数参数是否一致，还有是否缺少引用等细节

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