LeetCode-139-Word Break

Posted 2021-02-08 无名路人甲

算法描述：

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

解题思路：动态规划，递推式：

dp[i] = dp[j] && inDict(s.substr(j,i-j)) || dp[j-1] && inDict(s.substr(j-1,i-j+1)) || ...|| dp[0] && inDict(s.substr(0,i))

dp[0] = 0

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