[leetcode] 139. Word Break

Word Break

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.
Note:
The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.
Example 1:

Input: s = “leetcode”, wordDict = [“leet”, “code”]
Output: true
Explanation: Return true because “leetcode” can be segmented as “leet code”.

Example 2:

Input: s = “applepenapple”, wordDict = [“apple”, “pen”]
Output: true
Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”.
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = “catsandog”, wordDict = [“cats”, “dog”, “sand”, “and”, “cat”]
Output: false

解法1：动态规划

定义一个dp数组，大小为s.size()+1，初始化dp[0]=1，dp[i]表示s[0,i)数组可以拆分
用j把s[0,i)分为[0,j)和[j,i)两部分，dp[j]表示前一段是否能拆分，只需要判断后一段是否在字典中，就可以得到dp[i]是否能拆分。

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(),wordDict.end());
        vector<int> memo(s.size(),-1);
        vector<int> dp(s.size()+1,0);
        dp[0] = 1;
        for (int i = 0; i < dp.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                if (dp[j] && wordSet.count(s.substr(j, i - j))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp.back();
    }
};

解法2：DFS

记忆数组 memo[i] 定义为范围为 [i, n] 的子字符串是否可以拆分，初始化为 -1，表示没有计算过，如果可以拆分，则赋值为1，反之为0。

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(),wordDict.end());
        vector<int> memo(s.size(),-1);
        return helper(s,wordSet, memo, 0);
    }
    bool helper(string s, unordered_set<string>& wordSet, vector<int>& memo, int start){
        if(start == s.size())
            return true;
        if (memo[start] != -1) return memo[start];
        for(int i=start+1;i<=s.size();i++){
            string str = s.substr(start,i-start);
            if(wordSet.count(str) && helper(s, wordSet,memo,i)){
                return memo[start] = 1;
            }
        }
        return memo[start] = 0;
    }
};

解法3：BFS

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(),wordDict.end());
        vector<bool> visited(s.size());
        queue<int> q;
        q.push(0);
        while(!q.empty()){
            int t = q.front();
            q.pop();
            if (!visited[t]) {
                for(int i=t+1;i<=s.size();i++){
                    if(wordSet.count(s.substr(t,i-t))){
                        q.push(i);
                        if(i == s.size())
                            return true;
                    }  
                }
                visited[t] = 1;
            }
        }
        return false;
    }
};

参考

https://www.cnblogs.com/grandyang/p/4257740.html