[LeetCode 解题报告]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
class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int len = s.size();
        vector<int> res(len+1, false);
        res[0] = true;
        
        unordered_set<string> ss(wordDict.begin(), wordDict.end());l
        
        for (int i = 0; i < len + 1; i ++) {
            for (int j = 0; j < i; j ++) {
                if (res[j] && ss.find(s.substr(j, i-j)) != ss.end()) {
                    res[i] = true;
                    break;
                }
            }
        }
        return res[len];
    }
};
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转载自blog.csdn.net/caicaiatnbu/article/details/103838912