字符串编辑

阶乘除法

压缩字符串

<pre><code>s = &#x27;&#x27; lena = len(a) for i in range(lena): count = 1 t = i + 1 while t != lena and a[t] == a[i]: count += 1 t += 1 if a[i] != a[i-1] or i == 0: s += a[i] + str(count) lens = len(s) print(a) if lens &gt; lena else print(s) </code></pre>

就循环呗

ls=[] i=0 while i&lt;len(a):     j=i     cnt=0     ls.append(a[i])     while a[i]==a[j]:         i=i+1         cnt=cnt+1 if i==len(a): break     ls.append(cnt) if len(ls)&gt;len(a): print(a) else: for i in ls: print(i,end=&#x27;&#x27;)

蹦叉叉

<pre><code>a+=&quot;$&quot; #循环不到最后，给字符串尾部加一个不可能重复的字符 s=&#x27;&#x27; left=0 lens=len(a) for i in range(lens-1): if a[i] != a[i+1]: s+=a[i]+str(i+1-left) left=i+1 print(a[:-1] if len(s)&gt;lens else s)</code></pre>

虽然通过，感觉这个解法有点傻

<pre><code>def get_res(string): &quot;&quot;&quot; (想想成斜率，就比较容易理解) &quot;&quot;&quot;     tmp=&#x27;&#x27;     ind_start=0     string=string+&#x27;0&#x27;#结束符 for i in range(len(string)-1): if string[i]!=string[i+1]: # print(i+1,string[i+1])             sym=string[i]             tmp+=string[i]+str(string[ind_start:i+1].count(sym))             ind_start=i+1#更新起点     res=string[:-1] if len(string)-1&lt;len(tmp) else tmp return res print(get_res(a)) </code></pre>

For循环

re='' num=1 for i in range(len(a)-1): &nbsp;&nbsp;&nbsp; if num==1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; re=re+a[i] &nbsp;&nbsp;&nbsp; if a[i]==a[i+1]: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; num+=1 &nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; re+=str(num) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; num=1 if a[-1]==a[-2]: &nbsp;&nbsp;&nbsp; re+=str(num) else: &nbsp;&nbsp;&nbsp; re+=str(num) &nbsp;&nbsp;&nbsp; re+=a[-1] if len(re)&gt;len(a): &nbsp;&nbsp;&nbsp; print a else: &nbsp;&nbsp;&nbsp; print re

朴实解法

a=list(a) b=[] i=0 while i &lt;len(a): &nbsp; &nbsp; b.append(a[i]) &nbsp; &nbsp; cout=1 &nbsp; &nbsp; for j in range(i+1,len(a)): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if a[i]==a[j]: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; cout=cout+1 &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break &nbsp; &nbsp; b.append(str(cout)) &nbsp; &nbsp; i=cout+i if len(a)&lt;len(b): &nbsp; &nbsp; print(''.join(a)) else: &nbsp; &nbsp; print(''.join(b))

笨办法

<pre style="font-family: 宋体; font-size: 12pt;">a = 'aabcccccaaa'
l = {i:a.count(i) for i in a}
s = ''
for k,v in l.items():
 s += k + str(v)
 if len(s)&gt;len(a):
 print a
print s</pre>

第二步错了，看题目要求：例如：a="aabcccccaaa", 则输出：a2b1c5a3

在pc无报错简洁代码

<pre class="prettyprint lang-py linenums">same = 1
r = []
for i in range(len(a)-1):
 if a[i+1] == a[i]:
 same += 1
 else:
 r.append(a[i])
 r.append(str(same))
 same = 1
r.append(a[i+1])
r.append(str(same))
e = &quot;&quot;.join(r)
if len(a) &gt;= len(e):
 print(e)
else:
 print(a)</pre>

通过了

import re s = re.sub(r'(\w)(\1*)',lambda m:m.group(1)+str(1+len(m.group(2))), a) s = min(s, a, key=len) print(s)

re最简单了

b=list(a) c=[] k=1 i=1 try: &nbsp; for i in range(len(b)-1): &nbsp; &nbsp; if b[i]==b[i+1]: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;k+=1 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; c.append(b[i]) &nbsp; &nbsp; &nbsp; &nbsp; c.append(k) &nbsp; &nbsp; &nbsp; &nbsp; k=1 &nbsp; c.append(b[i]) &nbsp; c.append(k) except IndexError: &nbsp; &nbsp; print(&quot;list index out of range&quot;) if len(c)&gt;len(a): &nbsp; &nbsp; print(a) else: &nbsp; &nbsp; s=str(c) &nbsp; &nbsp; print(s)

right&do not run

这个代码比较直

输出是&nbsp;a2b1c5a&nbsp; 少了最后一个数字

l = a[0] i = 1 for j in range(0,len(a)-1): &nbsp; &nbsp; if a[j] != a[j+1]: &nbsp; &nbsp; &nbsp; &nbsp; l = l + str(i) &nbsp; &nbsp; &nbsp; &nbsp; l = l + a[j+1] &nbsp; &nbsp; &nbsp; &nbsp; i = 1 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; i = i + 1 if a[len(a) - 1] != a[len(a) - 2]: &nbsp; &nbsp; l = l + str(i) print(l)

第161题：压缩字符串

字符串编辑

计算GPA

放棋子

判断三角形