[LeetCode-Algorithms-139] "Word Break" (2017.12.14-WEEK15)

题目链接：Word Break

• 题目描述：
  

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”.

---

(1)思路：我们可以一个字符一个字符的判断，用一个bool类型的向量作为标识符，长度为字符串s的长度加1，每一位初始为false，第一位设为true。遍历字符串，如果该字符在dict中就将对应位置的标识符设为true，每一位的判断先要满足前一位为true的条件，如果字符串s长度的位置为true即代表可以完全分割。

(2)代码：

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