Question

139. 单词拆分

English Version

题目描述

给你一个字符串 s 和一个字符串列表 wordDict 作为字典。如果可以利用字典中出现的一个或多个单词拼接出 s 则返回 true。

注意：不要求字典中出现的单词全部都使用，并且字典中的单词可以重复使用。

 

示例 1：

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。

示例 2：

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。

示例 3：

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

 

提示：

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s 和 wordDict[i] 仅由小写英文字母组成
• wordDict 中的所有字符串 互不相同

解法

方法一：哈希表 + 动态规划

我们定义 $f[i]$ 表示字符串 $s$ 的前 $i$ 个字符能否拆分成 $wordDict$ 中的单词，初始时 $f[0]=true$，其余为 $false$。答案为 $f[n]$。

考虑 $f[i]$，如果存在 $j \in [0, i)$ 使得 $f[j] \land s[j:i] \in wordDict$，则 $f[i]=true$。为了优化效率，我们可以使用哈希表存储 $wordDict$ 中的单词，这样可以快速判断 $s[j:i]$ 是否在 $wordDict$ 中。

时间复杂度 $O(n^3)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。

class Solution:
  def wordBreak(self, s: str, wordDict: List[str]) -> bool:
    words = set(wordDict)
    n = len(s)
    f = [True] + [False] * n
    for i in range(1, n + 1):
      f[i] = any(f[j] and s[j:i] in words for j in range(i))
    return f[n]

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

class Solution {
public:
  bool wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> words(wordDict.begin(), wordDict.end());
    int n = s.size();
    bool f[n + 1];
    memset(f, false, sizeof(f));
    f[0] = true;
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j < i; ++j) {
        if (f[j] && words.count(s.substr(j, i - j))) {
          f[i] = true;
          break;
        }
      }
    }
    return f[n];
  }
};

func wordBreak(s string, wordDict []string) bool {
  words := map[string]bool{}
  for _, w := range wordDict {
    words[w] = true
  }
  n := len(s)
  f := make([]bool, n+1)
  f[0] = true
  for i := 1; i <= n; i++ {
    for j := 0; j < i; j++ {
      if f[j] && words[s[j:i]] {
        f[i] = true
        break
      }
    }
  }
  return f[n]
}

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

impl Solution {
  pub fn word_break(s: String, word_dict: Vec<String>) -> bool {
    let words: std::collections::HashSet<String> = word_dict.into_iter().collect();
    let mut f = vec![false; s.len() + 1];
    f[0] = true;
    for i in 1..=s.len() {
      for j in 0..i {
        f[i] |= f[j] && words.contains(&s[j..i]);
      }
    }
    f[s.len()]
  }
}

public class Solution {
  public bool WordBreak(string s, IList<string> wordDict) {
    var words = new HashSet<string>(wordDict);
    int n = s.Length;
    var f = new bool[n + 1];
    f[0] = true;
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j < i; ++j) {
        if (f[j] && words.Contains(s.Substring(j, i - j))) {
          f[i] = true;
          break;
        }
      }
    }
    return f[n];
  }
}

方法二：前缀树 + 动态规划

我们先将 $wordDict$ 中的单词存入前缀树中，然后使用动态规划求解。

我们定义 $f[i]$ 表示从字符串 $s$ 的第 $i$ 个字符开始往后拆分，能否拆分成 $wordDict$ 中的单词，初始时 $f[n]=true$，其余为 $false$。答案为 $f[0]$。

接下来，我们从大到小枚举 $i$，对于每个 $i$，我们从 $i$ 开始往后拆分，如果 $s[i:j]$ 在前缀树中，且 $f[j+1]=true$，则 $f[i]=true$。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。

class Trie:
  def __init__(self):
    self.children: List[Trie | None] = [None] * 26
    self.isEnd = False

  def insert(self, w: str):
    node = self
    for c in w:
      idx = ord(c) - ord('a')
      if not node.children[idx]:
        node.children[idx] = Trie()
      node = node.children[idx]
    node.isEnd = True


class Solution:
  def wordBreak(self, s: str, wordDict: List[str]) -> bool:
    trie = Trie()
    for w in wordDict:
      trie.insert(w)
    n = len(s)
    f = [False] * (n + 1)
    f[n] = True
    for i in range(n - 1, -1, -1):
      node = trie
      for j in range(i, n):
        idx = ord(s[j]) - ord('a')
        if not node.children[idx]:
          break
        node = node.children[idx]
        if node.isEnd and f[j + 1]:
          f[i] = True
          break
    return f[0]

class Solution {
  public boolean wordBreak(String s, List<String> wordDict) {
    Trie trie = new Trie();
    for (String w : wordDict) {
      trie.insert(w);
    }
    int n = s.length();
    boolean[] f = new boolean[n + 1];
    f[n] = true;
    for (int i = n - 1; i >= 0; --i) {
      Trie node = trie;
      for (int j = i; j < n; ++j) {
        int k = s.charAt(j) - 'a';
        if (node.children[k] == null) {
          break;
        }
        node = node.children[k];
        if (node.isEnd && f[j + 1]) {
          f[i] = true;
          break;
        }
      }
    }
    return f[0];
  }
}

class Trie {
  Trie[] children = new Trie[26];
  boolean isEnd = false;

  public void insert(String w) {
    Trie node = this;
    for (int i = 0; i < w.length(); ++i) {
      int j = w.charAt(i) - 'a';
      if (node.children[j] == null) {
        node.children[j] = new Trie();
      }
      node = node.children[j];
    }
    node.isEnd = true;
  }
}

class Trie {
public:
  vector<Trie*> children;
  bool isEnd;
  Trie()
    : children(26)
    , isEnd(false) {}

  void insert(string word) {
    Trie* node = this;
    for (char c : word) {
      c -= 'a';
      if (!node->children[c]) node->children[c] = new Trie();
      node = node->children[c];
    }
    node->isEnd = true;
  }
};

class Solution {
public:
  bool wordBreak(string s, vector<string>& wordDict) {
    Trie trie;
    for (auto& w : wordDict) {
      trie.insert(w);
    }
    int n = s.size();
    vector<bool> f(n + 1);
    f[n] = true;
    for (int i = n - 1; ~i; --i) {
      Trie* node = &trie;
      for (int j = i; j < n; ++j) {
        int k = s[j] - 'a';
        if (!node->children[k]) {
          break;
        }
        node = node->children[k];
        if (node->isEnd && f[j + 1]) {
          f[i] = true;
          break;
        }
      }
    }
    return f[0];
  }
};

type trie struct {
  children [26]*trie
  isEnd  bool
}

func newTrie() *trie {
  return &trie{}
}

func (t *trie) insert(w string) {
  node := t
  for _, c := range w {
    c -= 'a'
    if node.children[c] == nil {
      node.children[c] = newTrie()
    }
    node = node.children[c]
  }
  node.isEnd = true
}

func wordBreak(s string, wordDict []string) bool {
  trie := newTrie()
  for _, w := range wordDict {
    trie.insert(w)
  }
  n := len(s)
  f := make([]bool, n+1)
  f[n] = true
  for i := n - 1; i >= 0; i-- {
    node := trie
    for j := i; j < n; j++ {
      k := s[j] - 'a'
      if node.children[k] == nil {
        break
      }
      node = node.children[k]
      if node.isEnd && f[j+1] {
        f[i] = true
        break
      }
    }
  }
  return f[0]
}

function wordBreak(s: string, wordDict: string[]): boolean {
  const trie = new Trie();
  for (const w of wordDict) {
    trie.insert(w);
  }
  const n = s.length;
  const f: boolean[] = new Array(n + 1).fill(false);
  f[n] = true;
  for (let i = n - 1; i >= 0; --i) {
    let node: Trie = trie;
    for (let j = i; j < n; ++j) {
      const k = s.charCodeAt(j) - 97;
      if (!node.children[k]) {
        break;
      }
      node = node.children[k];
      if (node.isEnd && f[j + 1]) {
        f[i] = true;
        break;
      }
    }
  }
  return f[0];
}

class Trie {
  children: Trie[];
  isEnd: boolean;

  constructor() {
    this.children = new Array(26);
    this.isEnd = false;
  }

  insert(w: string): void {
    let node: Trie = this;
    for (const c of w) {
      const i = c.charCodeAt(0) - 97;
      if (!node.children[i]) {
        node.children[i] = new Trie();
      }
      node = node.children[i];
    }
    node.isEnd = true;
  }
}

public class Solution {
  public bool WordBreak(string s, IList<string> wordDict) {
    Trie trie = new Trie();
    foreach (string w in wordDict) {
      trie.Insert(w);
    }
    int n = s.Length;
    bool[] f = new bool[n + 1];
    f[n] = true;
    for (int i = n - 1; i >= 0; --i) {
      Trie node = trie;
      for (int j = i; j < n; ++j) {
        int k = s[j] - 'a';
        if (node.Children[k] == null) {
          break;
        }
        node = node.Children[k];
        if (node.IsEnd && f[j + 1]) {
          f[i] = true;
          break;
        }
      }
    }
    return f[0];
  }
}

class Trie {
  public Trie[] Children { get; set; }
  public bool IsEnd { get; set; }

  public Trie() {
    Children = new Trie[26];
    IsEnd = false;
  }

  public void Insert(string word) {
    Trie node = this;
    foreach (char c in word) {
      int i = c - 'a';
      if (node.Children[i] == null) {
        node.Children[i] = new Trie();
      }
      node = node.Children[i];
    }
    node.IsEnd = true;
  }
}