【常见题型总结】序列 DP 模板题（总结「线性 DP」和「序列 DP」本质区别）

输入: s = "leetcode", wordDict = ["leet", "code"]

输出: true

解释: 返回 true 因为 "leetcode" 可以由 "leet" 和 "code" 拼接成。

输入: s = "applepenapple", wordDict = ["apple", "pen"]

输出: true

解释: 返回 true 因为 "applepenapple" 可以由 "apple" "pen" "apple" 拼接成。
     注意，你可以重复使用字典中的单词。

输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]

输出: false

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> set = new HashSet<>();
        for (String word : wordDict) set.add(word);
        int n = s.length();
        boolean[] f = new boolean[n + 10];
        f[0] = true;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= i && !f[i]; j++) {
                String sub = s.substring(j - 1, i);
                if (set.contains(sub)) f[i] = f[j - 1]; 
            }
        }
        return f[n];
    }
}

function wordBreak(s: string, wordDict: string[]): boolean {
    const ss = new Set<string>(wordDict)
    const n = s.length
    const f = new Array<boolean>(n + 10).fill(false)
    f[0] = true
    for (let i = 1; i <= n; i++) {
        for (let j = i; j >= 1 && !f[i]; j--) {
            const sub = s.substring(j - 1, i)
            if (ss.has(sub)) f[i] = f[j - 1]
        }
    }
    return f[n]
}

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        ss = set(wordDict)
        n = len(s)
        f = [False] * (n + 10)
        f[0] = True
        for i in range(1, n + 1):
            j = i
            while j >= 1 and not f[i]:
                sub = s[j - 1:i]
                if sub in ss:
                    f[i] = f[j - 1]
                j -= 1
        return f[n]