README_EN.md

139. Word Break

中文文档

Description

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

 

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

 

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

Solutions

Dynamic Programming.

Python3

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        words = set(wordDict)
        n = len(s)
        dp = [False] * (n + 1)
        dp[0] = True
        for i in range(1, n + 1):
            for j in range(i):
                if dp[j] and s[j:i] in words:
                    dp[i] = True
                    break
        return dp[n]

Java

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

C++

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> words;
        for (auto word : wordDict) {
            words.insert(word);
        }
        int n = s.size();
        vector<bool> dp(n + 1, false);
        dp[0] = true;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (dp[j] && words.find(s.substr(j, i - j)) != words.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[n];
    }
};

Go

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

C#

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

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