139.Word Break

tags: Medium Array Hash Table String DP

139. Word Break

題目描述

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.

範例

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.

解答

C++

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = s.size();
        vector<bool> segmentable(n + 1, 0); // s[0:ed] can be segmented into wordDict
        segmentable[0] = true;
        for (int ed = 1; ed <= n; ed ++) {
            for (int st = 0; st < ed; st ++) {
                string testStr = s.substr(st, ed - st);
                bool found = find(wordDict.begin(), wordDict.end(), testStr) != wordDict.end();
                if (segmentable[st] and found) {
                    segmentable[ed] = true;
                    break;
                }
            }
        }
        return segmentable[n];
    }
};

Jerry Wu4 August, 2023

Python

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        n = len(s)
        @cache
        def bt(i):
            if i == n:
                return True
            return any(bt(j+1) for j in range(i, n) if s[i: j+1] in wordDict)
        return bt(0)

Yen-Chi ChenSat, Aug 5, 2023

Reference

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