LeetCode_Everyday：017 Letter Combinations of a Phone Number

LeetCode Everyday：堅持價值投資，做時間的朋友！！！

題目:

給定一個僅包含數字2-9的字符串，返回所有它能表示的字母組合。
給出數字到字母的映射如下（與電話按鍵相同）。注意 1 不對應任何字母。

示例：

• 示例 1:

  輸入："23"
  輸出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
  

• 說明：

  儘管上面的答案是按字典序排列的，但是你可以任意選擇答案輸出的順序。

代碼

方法一： python中嵌套for循環

執行用時：44 ms, 在所有 Python3 提交中擊敗了40.93%的用戶
內存消耗：13.8 MB, 在所有 Python3 提交中擊敗了5.41%的用戶

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        conversion={'2':'abc','3':'def','4':'ghi','5':'jkl','6':'mno','7':'pqrs','8':'tuv','9':'wxyz'}
        if len(digits)==0:
            return [] 
        product=['']
        for k in digits:
            product=[i+j for i in product for j in conversion[k]]
        return product

"""
For Example:    input:   digits = "23"
               output:   ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
"""
digits = "23"
                
solution = Solution()
result = solution.letterCombinations(digits)
print('輸出爲:', result)

方法二： 回溯

執行用時：40 ms, 在所有 Python3 提交中擊敗了65.51%的用戶
內存消耗：13.5 MB, 在所有 Python3 提交中擊敗了5.41%的用戶

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if not digits: return []

        phone = {'2':['a','b','c'],
                 '3':['d','e','f'],
                 '4':['g','h','i'],
                 '5':['j','k','l'],
                 '6':['m','n','o'],
                 '7':['p','q','r','s'],
                 '8':['t','u','v'],
                 '9':['w','x','y','z']}
                
        def backtrack(conbination,nextdigit):
            if len(nextdigit) == 0:
                res.append(conbination)
            else:
                for letter in phone[nextdigit[0]]:
                    backtrack(conbination + letter,nextdigit[1:])

        res = []
        backtrack('',digits)
        return res
    
"""
For Example:    input:   digits = "23"
               output:   ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
"""
digits = "23"
                
solution = Solution()
result = solution.letterCombinations(digits)
print('輸出爲:', result)

方法三： 隊列

執行用時：40 ms, 在所有 Python3 提交中擊敗了65.51%的用戶
內存消耗：13.8 MB, 在所有 Python3 提交中擊敗了5.41%的用戶

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if not digits: return []
        phone = ['abc','def','ghi','jkl','mno','pqrs','tuv','wxyz']
        queue = ['']  # 初始化隊列
        for digit in digits:
            for _ in range(len(queue)):
                tmp = queue.pop(0)
                for letter in phone[ord(digit)-50]:# 這裏我們不使用 int() 轉換字符串，使用ASCII碼
                    queue.append(tmp + letter)
        return queue

"""
For Example:    input:   digits = "23"
               output:   ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
"""
digits = "23"
                
solution = Solution()
result = solution.letterCombinations(digits)
print('輸出爲:', result)

參考

1. https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/solution/pythondian-hua-hao-ma-de-zi-mu-zu-he-by-jutraman/
2. https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/solution/hui-su-dui-lie-tu-jie-by-ml-zimingmeng/

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