LeetCode 139. Word Break

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. 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.

/*题意：给定字符串s, 问是否能够由字典dict中的单词组成思路：dp[j]表示前j个字符可以由字典dict中的单词组成,对于长度为i的字符串，若dp[j]为真且子串s.substr(j, i-j)存在于字典dict中，则长度为i的字符串能够由字典dict中的单词组成（j >= 0 && j < i）*/class Solution {public:    bool find(vector<string>& wordDict, string s){        //用来判断字符串s是否在字典dict中        for(int k = 0; k < wordDict.size(); k++){            if(s == wordDict[k])                return true;        }        return false;    }    bool wordBreak(string s, vector<string>& wordDict) {        vector<bool>dp(s.length() + 1, false);        dp[0] = true;        for(int i = 1; i < s.length() + 1; i++){            for(int j = i - 1; j >= 0; j--){                if(dp[j] && find(wordDict, s.substr(j, i-j))){                    //只要找到一种切分方式就说明长度为i的字符串可以成功切分                    dp[i] = true;                    break;                }            }        }        return dp[s.length()];    }};