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leetcode(140) - 单词拆分II

2019/01/01

leetcode(140) - 单词拆分II

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, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

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 = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

2 中文版

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

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 = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

二、ruby方案

2.1 思路

需要返回所有结果，那么回溯基本上跑不了了。

不过这道题不能强行回溯，会超时的，需要用到上一题的动态规划先判断字符串能不能被拆分，如果可以再进行回溯。

2.2 代码实现

# @param {String} s
# @param {String[]} word_dict
# @return {String[]}
def word_break(s, word_dict)
  # 深搜
  split_flag = can_break?(s, word_dict)
  if split_flag[-1]
    word_dict = word_dict.uniq
    res_list, array = [], []
    return word_break_dfs(s, word_dict, res_list, array, split_flag, s.size, 0)
  else
    []
  end
end

def word_break_dfs(s, word_dict, res_list, array,  split_flag, length, start)
  if start == length - 1
    res_list << array
    return
  end

  start.upto(length - 1).each do |index|
    str = s[start..index]
    if word_dict.include? str
      array << str
      word_break_dfs(s, word_dict, res_list, array,  split_flag, length, index)
    end
  end
end

def can_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
end


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