LeetCode | 139. Word Break

139. Word Break

Link: https://leetcode.com/problems/word-break/

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.

Approach

  • Initialize a boolean dp array to keep track of whether the substring ending at index i can be segmented into words from the wordDict.
  • Create a HashSet set with wordDict to check if a substring is a valid word.
  • Set dp[0] to true, indicating that an empty string can always be segmented.
  • Iterate through the string s using a loop with i ranging from 1 to the length of s. For each i, iterate through another loop with j ranging from 0 to i. The purpose of this nested loop is to explore different substrings within the string s.
    • Check if the substring from index j to index i is present in the set and if dp[j] is also true. If both conditions are met, it means that the substring from j to i can be segmented, so set dp[i] to true.

Solution

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> set = 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 && !dp[i]; j++) {
                if (set.contains(s.substring(j, i)) && dp[j])
                    dp[i] = true;
            }
        }
        return dp[s.length()];
    }
}

你可能感兴趣的:(leetcode,算法)