给定一个整数数组和一个目标值，找出数组中和为目标值的两个数。你可以假设每个输入只对应一种答案，且同样的元素不能被重复利用。
python：
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
n = len(s)
if n == 0:
return s
else:
maxNum = 1
start = 0
for i in range(n):
count = 1
left = i-1
right = i+1
while left>=0 and right < n:
if s[left] == s[right]:
count += 2
left -= 1
right += 1
else:
break
if count>=maxNum:
maxNum = count
start = left+1
for j in range(n):
count = 0
r = j
l = j-1
while l>=0 and r<n:
if s[l] == s[r]:
count += 2
l -= 1
r += 1
else:
break
if count>=maxNum:
maxNum = count
start = l+1
return s[start:(start+maxNum)]C++：
class Solution {
public:
string longestPalindrome(string s) {
int n=s.length();
if(n == 0){
return s;
}
else{
int start = 0;
int maxNum = 1;
for(int i=0;i<n;i++){
int left = i-1;
int right = i + 1;
int count = 1;
while(left>=0 && right<n){
if(s[left]==s[right]){
count += 2;
left -= 1;
right += 1;
}
else{
break;
}
}
if(count>maxNum){
maxNum = count;
start = left+1;
}
}
for(int i=0;i<n;i++){
int right = i;
int left = i-1;
int count = 0;
while(left>=0 && right<n){
if(s[left]==s[right]){
count += 2;
left -= 1;
right += 1;
}
else{
break;
}
}
if(count>maxNum){
maxNum = count;
start = left+1;
}
}
return s.substr(start,maxNum);
}
}
};