139. Word Break(图解)

139. 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

Solution

C++

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

Explanation
dp:
139. Word Break(图解)_第1张图片

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