【29】139. Word Break

139. Word Break

• Total Accepted: 127484
• Total Submissions: 447287
• Difficulty: Medium
• Contributors: Admin

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.

class Solution {
public:
//DP
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = s.size();
        unordered_set<string> dict;
        for(string ss : wordDict){
            dict.insert(ss);
        }
        vector<bool> dp(n + 1, false);
        dp[0] = true;//s为空的时候是true
        for(int i = 1; i < n + 1; i++){
            for(int j = 0; j < i; j++){
                if(dp[j] && dict.find(s.substr(j, i - j)) != dict.end()){//vector没有find方法，用set
                    dp[i] = true;
                    break;//只要找到一种解法 到第i个char时可以分解 就break
                }
                
            }
        }
        return dp[n];
    }
};