139. 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

思路：

Solution (C++):

bool wordBreak(string s, vector<string>& wordDict) {
    if (wordDict.empty())  return false;
    int n = s.size();
    vector<bool> dp(n+1, false);
    dp[0] = true;

    for (int i = 1; i <= n; ++i) {
        for (int j = i-1; j >= 0; --j) {
            if (dp[j]) {
                string word = s.substr(j, i-j);
                if (find(wordDict.begin(), wordDict.end(), word) != wordDict.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }
    }
    return dp[n];
}

性能：

Runtime: 4 ms  Memory Usage: 10.5 MB