leetcode:Word Break

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".

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
    
        set<int> dicWordLen;
        unordered_set<string>::iterator itr=dict.begin();
        
        while (itr != dict.end())
        {
            int len = (*itr).size();
            dicWordLen.insert(len);
            ++itr;
        }

        int strLen = s.size();
        vector<bool> D(strLen+1);
        
        for (int i=0; i<strLen+1; i++)
        {
            D[i] = false;
        }
        
        D[0] = true;//这个初始化很重要
        
        for (int i = 0; i < strLen; i++)
        {
            set<int>::iterator itr = dicWordLen.begin();
            
            while (itr != dicWordLen.end())
            {
                if ((*itr) <= i+1) //the len from 0 to i is i+1 在这里我们只考虑在字典里长度比当前长度小的那些词
                {
                    if (D[i+1-(*itr)] == true) 
                    {
                        string subString = s.substr(i-(*itr)+1, (*itr));
                        
                        unordered_set<string>::iterator itr = dict.find(subString);
                        if (itr != dict.end())
                        {
                            D[i+1] = true;
                            break;
                        }
                    }
                }
                else
                {
                    break;
                }
                ++itr;
            }
        }
        
        return D[strLen];
    }
};