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leetcode(139) - 单词拆分

2019/01/01

leetcode(139) - 单词拆分

一、题目描述

1 英文版

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

2 中文版

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict，判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词。

说明：

  • 拆分时可以重复使用字典中的单词。
  • 你可以假设字典中没有重复的单词。

示例 1：

输入: s = "leetcode", wordDict = ["leet", "code"]
输出: true
解释: 返回 true 因为 "leetcode" 可以被拆分成 "leet code"。

示例 2：

输入: s = "applepenapple", wordDict = ["apple", "pen"]
输出: true
解释: 返回 true 因为 "applepenapple" 可以被拆分成 "apple pen apple"。
     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

二、ruby方案

2.1 思路

动态规划，dp[i]表示[0,i)的字符串能否被分割成单词组合（true or false）

则 dp[i] = dp[j] && dict.include?(str[j..i]) 0 <= j < i

2.2 代码实现

# @param {String} s
# @param {String[]} word_dict
# @return {Boolean}
def word_break(s, word_dict)
  length = s.size

  split_flag = Array.new(length + 1, false)
  split_flag[0] = true

  1.upto(length).each do |i|
    0.upto(i - 1).each do |j|
      if split_flag[j] && word_dict.include?(s[j...i])
        split_flag[i] = true
        break
      end
    end
  end

  split_flag[-1]
end


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