139. Word Break

https://leetcode.com/problems/word-break/

Problem

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) {
        const int n = s.size() + 1;
        bool dp[n];
        memset(dp, false, sizeof(bool) * n);
        dp[0] = true;
        for (int i = 1; i < n; ++i) {
            for (const auto &str: wordDict) {
                const int len = str.size();
                if (i - len < 0) continue;
                bool match = true;
                for (int j = len - 1, k = 1; j >= 0; --j, ++k) {
                    if (str[j] != s[i - k]) {
                        match = false; break;
                    }
                }
                dp[i] = match && dp[i - len];
                if (dp[i]) break;
            }
        }
        return dp[n - 1];
    }
};

• #dp