LeetCode: 139. Word Break

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139. Word Break

LeetCode - Word Break

TODO: 似乎明白了，又似乎不明白。

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

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/**
 * Runtime: 11 ms, faster than 6.65% of Java online submissions for Word Break.
 *
 * Memory Usage: 44.3 MB, less than 5.08% of Java online submissions for Word Break.
 *
 * Copy from: https://leetcode-cn.com/problems/word-break/solution/dan-ci-chai-fen-by-leetcode/[单词拆分 - 单词拆分 - 力扣（LeetCode）]
 */
public boolean wordBreak(String s, List<String> wordDict) {
    boolean[] dp = new boolean[s.length() + 1];
    Arrays.fill(dp, false);
    dp[0] = true;
    Set<String> dict = new HashSet<>(wordDict);
    for (int i = 1; i <= s.length(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && dict.contains(s.substring(j, i))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.length()];
}