Longest Palindromic Substring
回文这种简单的问题,在C里面印象很深啊。希望能一次过。
写的时候才想到有两种情况:
454(奇数位)
4554(偶数位)
第1次提交
class Solution:
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
print(s)
maxPadStr=''
# traverse each char
l=len(s)
for i,c in enumerate(s):
# left or right extend test
# j is length
j=1
while j <= l/2:
# left or right str
leftStr=s[i:i-j:-1] if (i-j)>=0 else s[i::-1]
rightStr=s[i:i+j] # odd
rightStr2=s[i+1:i+j+1] # even
# length
leftLen=len(leftStr)
rightLen=len(rightStr)
print(i,j,'left:',leftStr,'right:',rightStr,'right-even:',rightStr2)
if leftLen != rightLen:
break
if leftStr == rightStr and len(leftStr)*2-1>=len(maxPadStr):
#odd bit
maxPadStr=(leftStr[:0:-1]+rightStr) # 654,654 -> 45654
#print('odd',maxPadStr)
# 'cbbd' 时,left: b right: b right-even: b. odd even 都存在,但上面只会b,所以存在偶数位相等故不能elif
if leftStr == rightStr2 and len(leftStr)*2>=len(maxPadStr):
#even bit
maxPadStr=(leftStr[::-1]+rightStr2) # 654,654 -> 456654
#print('even',maxPadStr)
j+=1
return maxPadStr
if __name__ == "__main__":
data = [
{
"input":"babad",
"output":"bab", # /'aba'
},{
"input":"cbbd",
"output":"bb"
},{
"input":"123456654",
"output":"456654"
},{
"input":"14565422",
"output":"45654"
}
];
for d in data:
result=Solution().longestPalindrome(d['input'])
print(result)
if result==d['output']:
print("--- ok ---")
else:
print("--- error ---")
Wrong Answer:
Input:
"a"
Output:
""
Expected:
"a"
咦,这种一个字母早都考虑到了啊。
出在了while的条件上:j <= l/2:
哈哈,一半长度的直观出发哈哈哈l/2,没错,只是把1个字符就排除掉了,改成这样j<=l//2+1
。
第2次提交
pass
while j <= l//2+1:
pass
Accepted!