给定一个字符串 S,返回 “反转后的” 字符串,其中不是字母的字符都保留在原地,而所有字母的位置发生反转。
示例 1:
输入:"ab-cd" 输出:"dc-ba"
示例 2:
输入:"a-bC-dEf-ghIj" 输出:"j-Ih-gfE-dCba"
示例 3:
输入:"Test1ng-Leet=code-Q!" 输出:"Qedo1ct-eeLg=ntse-T!"
提示:
- S.length <= 100
- 33 <= S[i].ASCIIcode <= 122
- S 中不包含 or "
class Solution
{
public:
string reverseOnlyLetters(string S)
{
int len = S.size();
if(len <= 1)
return S;
int low = 0;
int high = len - 1;
while(low < high)
{
for(; low < high; low++)
{
if(S[low] >= 'a' && S[low] <= 'z' || S[low] >= 'A' && S[low] <= 'Z')
{
break;
}
}
for(; high > low; high--)
{
if(S[high] >= 'a' && S[high] <= 'z' || S[high] >= 'A' && S[high] <= 'Z')
{
break;
}
}
if(low < high)
{
swap(S[low], S[high]);
low++;
high--;
}
}
return S;
}
};