[LeetCode] 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

题意，给一个字符串，给一个字符串数组，问字符数组能否组装出字符串
我佛了，这题，先说几个注意的地方，字符数组里的元素是可以重复使用的，字符数组里的元素是可以有剩余的
一开始自信满满的用set直接解，每个元素直接扫，被“aaaaaaa”，【“aaa”，“aaaa”】 卡死了
最后看了别人的博客才恍然大悟，挖槽，原来是到DP题
真的是醉了，LeetCode上DP题型实在是难受
关键在于拆的地方，直接上代码

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> set = new HashSet<>();
        for (int i = 0; i < wordDict.size(); i++) {
            set.add(wordDict.get(i));
        }
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 1; i < s.length() + 1; i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && set.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}