擅长:python、mysql、java
<p>可能的解决办法之一是</p>
<pre><code>inp=['w', 'o', 'w', 'w', 'y', 'y', 'o', 'w', 'w', 'o', 'y', 'o', 'w']
d=dict()
#store indices in dictionary
for i in range(len(inp)):
if inp[i]!='w':
if d.get(inp[i])!=None:
d[inp[i]].append(i)
else:
d[inp[i]]=[i]
# for 'w' in mid
for each in d:
l=d[each]
if len(l)>1:
for i in range(len(l)-1):
flag=0
for j in range(l[i]+1,l[i+1]):
flag=0
if inp[j]=='w':
flag=1
else:
flag=0
break
if flag==1:
for j in range(l[i]+1,l[i+1]):
inp[j]=each
#for 'w' at beginning
for i in range(len(inp)):
if inp[i]!='w':
for j in range(i):
inp[j]=inp[i]
break
#for 'w at end'
for i in range(len(inp)-1,-1,-1):
if inp[i]!='w':
for j in range(i+1,len(inp)):
inp[j]=inp[i]
break
print(inp)
</code></pre>