139. Word Break

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LeetCode

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Problem

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

The same word in the dictionary may be reused multiple times in the segmentation. You may assume the dictionary does not contain duplicate words. 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

Code

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> wordDictSet = new HashSet(wordDict);
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;

        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j < i; j++) {
                // 有一次成功就可以退出这次循环了
                if (dp[j] && wordDictSet.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp[s.length()];
    }
}

Time complexity : O(n^3) There are two nested loops, and substring computation at each iteration

Space complexity : O(n)

Recursion with memoization

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        return wordBreakMemo(s, new HashSet<>(wordDict), 0, new Boolean[s.length()]);
    }

    private boolean wordBreakMemo(
        String s,
        Set<String> wordDict,
        int start,
        Boolean[] memo) {
        if (start == s.length()) {
            return true;
        }
        if (memo[start] != null) {
            return memo[start];
        }
        for (int end = start + 1; end <= s.length(); end++) {
            if (wordDict.contains(s.substring(start, end)) && wordBreakMemo(s, wordDict, end, memo)) {
                memo[start] = true;
                return true;
            }
        }
        memo[start] = false;
        return false;
    }
}

Time complexity : O(2^n) Space complexity : O(2^n)