LeetCode 0139 - Word Break

Reck Zhang

发布于 2021-08-11 14:47:34

1770

发布于 2021-08-11 14:47:34

文章被收录于专栏：Reck Zhang

Word Break

Desicription

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

class Solution {
public:
    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 = i-1; j >= 0; j--) {
                if(dp[j] && dict.find(s.substr(j, i-j)) != dict.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.size()];
    }
};

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原始发表：2018-05-29，如有侵权请联系 cloudcommunity@tencent.com 删除

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