LeetCode Medium: 17. Letter Combinations of a Phone Number

 

一、题目

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example:

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

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

给定一个数字串，每一个数字所代表的的与电话号码上的一样，对每一个数字代表的字符进行组合。

二、思路

递归或者迭代，每迭代一哥 新的字符，就应该在上一次结果的基础上进行组合。下面给出三种不同的方式写的代码，思想都是一样。这些代码分别来自三个不同的博客，后面有附。

三、代码

#coding:utf-8
class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if digits == '':
            return []
        self.DigitDict=[' ','1', "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
        res = ['']
        for d in digits:
            res = self.letterCombBT(int(d),res)
        print(res)
        return res

    def letterCombBT(self, digit, oldStrList):
        return [dstr+i for i in self.DigitDict[digit] for dstr in oldStrList]

class Solution1(object):
    def letterCombinations1(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        digits = digits.strip()
        if len(digits)==0:
            return []
        rtn_queue=[""]
        head_ind=0  # the head of queue
        head_ele="" # the head element of queue
        mapping= ["0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
        dig_ind=0
        for dig in digits:
            x = int(dig)
            while len(rtn_queue[head_ind])==dig_ind:
                head_ele=rtn_queue[head_ind]
                head_ind+=1
                for ch in mapping[x]:
                    rtn_queue.append(head_ele+ch)
            dig_ind+=1
        queue_n = len(rtn_queue)
        print(rtn_queue[head_ind:queue_n])
        return rtn_queue[head_ind:queue_n]

class Solution2:
    map=[" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
    length=0;res=[]
    # @param {string} digits
    # @return {string[]}
    def letterCombinations2(self, digits):
        self.length=len(digits)
        self.res=[]
        if self.length==0:return self.res;
        tmp=['' for i in range(self.length)]
        self.getLetterCom(0,digits,tmp)
        print(self.res)
        return self.res
    def getLetterCom(self,index,digits,tmp):
        if index>=self.length:
            letters=''.join(tmp)
            self.res.append(letters)
            return
        digit=ord(digits[index])-ord('0')
        for i in range(len(self.map[digit])):
            tmp[index]=self.map[digit][i]
            self.getLetterCom(index+1,digits,tmp)



if __name__ == '__main__':
    import time
    a = "23"

    start = time.clock()
    ss = Solution()
    ss.letterCombinations(a)
    end = time.clock()
    cost = end - start
    print(cost)

    start1 = time.clock()
    ss1 = Solution1()
    ss1.letterCombinations1(a)
    end1 = time.clock()
    print(end1 - start1)

    ss2 = Solution2()
    ss2.letterCombinations2(a)

　　

参考博客：https://blog.csdn.net/zl87758539/article/details/51675166      https://blog.csdn.net/daigualu/article/details/74091229   https://blog.csdn.net/runningtortoises/article/details/45620177