[leetcode]Word Break

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".

方法有参考:动态规划

class Solution {public:    //动态规划的问题最后要从后向前去做，简单些    bool wordBreak(string s, unordered_set<string> &dict) {        size_t n = s.size();        if(n == 0) return true;                vector<bool> vb(n + 1, false);        vb[n] = true;                for(int i = n - 1; i >= 0; i--){            for(unordered_set<string>::iterator itor = dict.begin(); itor != dict.end(); itor++){                int dwordlen = itor->size(); //字典中单词的长度                int tlen = n - i; //当前要查找的字符串位置 到末尾的距离                if(tlen >= dwordlen && !vb[i]){                    //一定要保证在i + dwordlen之前的字符串可以查找到                    if(vb[i + dwordlen] && s.substr(i, dwordlen) == *itor){                         vb[i] = true; break;                    }                }            }        }                return vb[0];    }};