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中等
字典树
记忆化搜索
数组
哈希表
字符串
动态规划

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。

Python3

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]

Java

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];
    }
}

C++

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];
    }
};

Go

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]
}

TypeScript

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];
}

Rust

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()]
    }
}

C#

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。

Python3

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]

Java

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;
    }
}

C++

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];
    }
};

Go

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]
}

TypeScript

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;
    }
}

C#

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;
    }
}