Leetcode题解——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".

解题思路：这种字符串匹配的问题一般都是用动态规划或者递归解决。

class Solution {public:    bool wordBreak(string s, unordered_set<string>& wordDict) {        int n = s.size();        if(!n) return true;        if(wordDict.empty()) return false;                vector<bool> D(n+1,false);        D[0] = true;                int longestWord = 0;        for(auto word : wordDict){            longestWord = max(longestWord, (int)word.size());        }                for(int i=1; i<=n; i++){            int k = max(0,(i-longestWord));            for(int j=i-1; j>=k; j--){                if(D[j]){                    string t = s.substr(j,i-j);                    if(wordDict.find(t)!=wordDict.end()){                        D[i] = true;                        break;                    }                }            }        }        return D[n];            } };

动态规划：0ms

class Solution {public:     bool wordBreak(string s, unordered_set<string>& wordDict) {        int n = s.size();        if(!n) return true;        if(wordDict.empty()) return false;        for(int i=n-1; i>=0; i--){            if(wordDict.find(s.substr(i))!=wordDict.end()) break;            if(i==0) return false;        }        return findwordBreak(s,wordDict);    }        bool findwordBreak(string s, unordered_set<string>& wordDict){        int n = s.size();        if(n==0){            return true;        }        for(int i=1; i<=n; i++){            string t = s.substr(0,i);            if(wordDict.find(t)!=wordDict.end()){                if(findwordBreak(s.substr(i), wordDict)) return true;            }        }        return false;    }};

递归：0ms