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一行Python代码——电话簿上的字符.html
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一行Python代码——电话簿上的字符.html
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<meta property="og:description" content="Question¶ Given a digit string, 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. Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. Answer¶ num2letter = { '1':'', '2':'abc', '3 ..." />
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<meta name="twitter:description" content="Question¶ Given a digit string, 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. Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. Answer¶ num2letter = { '1':'', '2':'abc', '3 ...">
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<h1><a href="https://chukeer.github.io/一行Python代码——电话簿上的字符.html"> 一行Python代码——电话簿上的字符 </a></h1>
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<h3 id="question">Question<a class="headerlink" href="#question" title="Permanent link">¶</a></h3>
<p>Given a digit string, return all possible letter combinations that the number could represent.</p>
<p>A mapping of digit to letters (just like on the telephone buttons) is given below.</p>
<p><img alt="" src="http://littlewhite.us/pic/20150512/1.png"/></p>
<div class="highlight"><pre><span></span><span class="n">Input</span><span class="o">:</span><span class="n">Digit</span> <span class="n">string</span> <span class="s2">"23"</span>
<span class="n">Output</span><span class="o">:</span> <span class="o">[</span><span class="s2">"ad"</span><span class="o">,</span> <span class="s2">"ae"</span><span class="o">,</span> <span class="s2">"af"</span><span class="o">,</span> <span class="s2">"bd"</span><span class="o">,</span> <span class="s2">"be"</span><span class="o">,</span> <span class="s2">"bf"</span><span class="o">,</span> <span class="s2">"cd"</span><span class="o">,</span> <span class="s2">"ce"</span><span class="o">,</span> <span class="s2">"cf"</span><span class="o">].</span>
</pre></div>
<h3 id="answer">Answer<a class="headerlink" href="#answer" title="Permanent link">¶</a></h3>
<div class="highlight"><pre><span></span>num2letter = {
'1':'',
'2':'abc',
'3':'def',
'4':'ghi',
'5':'jkl',
'6':'mno',
'7':'pqrs',
'8':'tuv',
'9':'wxyz'}
def letterCombinations(digits):
return [] if digits == "" else reduce(lambda x, y:reduce(lambda a,b:a+b, [[i+j for i in x] for j in y], []), filter(lambda x:x!='', [num2letter[digit] for digit in digits]), [''])
</pre></div>
<h3 id="_1">答案分析<a class="headerlink" href="#_1" title="Permanent link">¶</a></h3>
<p>我们先求得每个数字对应的字符串，然后求这些字符串能组合的所有情况，比如根据数字得到的字符为'abc', 'def','ghi',那么组合数是3<em>3</em>3=27种，具体思路就是先求'abc'和'def'的组合['ab','ae','af','bd','be','bf','cd','ce','cf']，再算和'ghi'的组合，以此类推，这本身是每什么难度的，在Python里正好有个内建函数是干这个事的，那就是reduce</p>
<p>reduce的函数原型如下</p>
<div class="highlight"><pre><span></span>def reduce(function, iterable, initializer=None):
it = iter(iterable)
if initializer is None:
try:
initializer = next(it)
except StopIteration:
raise TypeError('reduce() of empty sequence with no initial value')
accum_value = initializer
for x in it:
accum_value = function(accum_value, x)
return accum_value
</pre></div>
<p>我们举个例子来说明</p>
<div class="highlight"><pre><span></span>reduce(lambda a,b:a+b, [1,2,3], 100)
</pre></div>
<p>这里返回的是100+1+2+3，lambda a,b:a+b定义了一个匿名函数，等价于</p>
<div class="highlight"><pre><span></span>def add(a, b):
return a + b
</pre></div>
<p>这个reduce函数的具体计算步骤如下</p>
<div class="highlight"><pre><span></span>1. redult = 100
2. result = add(result, 1) # 101
3. result = add(result, 2) # 103
4. result = add(result, 3) # 106*
5. return result
</pre></div>
<p>现在再来分析答案中的letterCombinations函数</p>
<p>为了看得清晰，我们先使用一些中间变量，改写函数如下</p>
<div class="highlight"><pre><span></span>def letterCombinations(digits):
data = filter(lambda x:x!='', [num2letter[digit] for digit in digits])
return [] if digits == "" else reduce(lambda x, y:reduce(lambda a,b:a+b, [[i+j for i in x] for j in y], []), data, [''])
</pre></div>
<ul>
<li>
<p>[num2letter[digit] for digit in digits]</p>
<p>根据digits获取对应字符串</p>
</li>
<li>
<p>filter(lambda x:x!='', [num2letter[digit] for digit in digits])</p>
<p>过滤掉空字符串</p>
</li>
<li>
<p>return [] if digits == "" else something</p>
<p>python的三元组表达式，类似C的<code>condition ? value1 : value2</code>，Python的形式为<code>value1 if condition else value2</code>，这里意思是如果digits为空则返回[]，否则返回后面的something，这里的something就是我们要重点分析的如下表达式</p>
<div class="highlight"><pre><span></span> reduce(lambda x, y:reduce(lambda a,b:a+b, [[i+j for i in x] for j in y], []), data, [''])
</pre></div>
</li>
</ul>
<p>下面我们来一层层抽丝剥茧。先看外层的reduce，我们将其表示成如下</p>
<div class="highlight"><pre><span></span>reduce(func(x,y), data, [''])
</pre></div>
<p>这是这道题的核心解题思路，我们还是以例子来说明，假设data=['abc','def','ghi']，那么计算步骤应该是</p>
<div class="highlight"><pre><span></span>1. result = ['']
2. result = func(result, 'abc')
3. result = func(result, 'def')
4. result = func(result, 'ghi')
5. return result
</pre></div>
<p>我们只需要实现func函数，问题就迎刃而解</p>
<p>其实func函数要做的就是将y里的每个字符添加到x里的每个字符串的末尾，比如</p>
<div class="highlight"><pre><span></span>x=['ab', 'cd']
y='ef'
</pre></div>
<p>那么func(x,y)应该返回<code>['abe','cde', 'abf', 'cdf']</code>，然后就是求func函数的实现</p>
<p>很容易我们想到列表解析，这里需要两层嵌套如下</p>
<div class="highlight"><pre><span></span>def func(x,y):
return [[i + j for i in x] for j in y]
</pre></div>
<p>但是这样的结果是[['abe', 'cde'], ['abf', 'cdf']]，此时我们还需要将列表的每个元素（注意，元素类型也为列表）连接起来，这不就是元素求和吗，只是这里的元素是列表而已，我们以文章开头reduce为参考</p>
<div class="highlight"><pre><span></span>reduce(lambda a,b:a+b, [1,2,3], 100)
</pre></div>
<p>只需稍作替换即可</p>
<div class="highlight"><pre><span></span>reduce(lambda a,b:a+b, [[i + j for i in x] for j in y], [])
</pre></div>
<p>这样func(x,y)函数的实现就成了下面这样</p>
<div class="highlight"><pre><span></span>def func(x,y):
return reduce(lambda a,b:a+b, [[i + j for i in x] for j in y, [])
</pre></div>
<p>我们再将最外层reduce中的func函数改写成lambda表达式</p>
<div class="highlight"><pre><span></span>reduce(lambda x,y:reduce(lambda a,b:a+b, [[i + j for i in x] for j in y], []), data, [''])
</pre></div>
<p>大功告成！</p>
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