Leetcode 139. Word Break

Tyan

发布于 2019-05-25 22:55:32

3260

发布于 2019-05-25 22:55:32

文章被收录于专栏：SnailTyanSnailTyan

文章作者：Tyan

博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

Version 1

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        if(s.size() == 0) {
            return false;
        }
        vector<int> flags(s.size(), -1);
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        return split(s, dict, flags, 0);
    }
    
    
    bool split(string& s, set<string>& dict, vector<int>& flags, int start) {
        if(start == s.size()) {
            return true;
        }
        if(flags[start] != -1) {
            return flags[start];
        }
        for(int i = start + 1; i <= s.size(); i++) {
            string temp = s.substr(start, i - start);
            if(dict.find(temp) != dict.end() && split(s, dict, flags, i)) {
                flags[start] = 1;
                return true;
            }
        }
        flags[start] = 0;
        return false;
    }
};

Version 2

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

Version 3

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