源代码:
//cpp.Leetcode OJ Contest23 Reverse String II_Leetcode541
/*大意:就是我有一串字符，每2k个字符，只把前K个反转，计剩下还有len个字符，如果剩下的字符 大于K小于2K，那么只反转前K个，剩下 len-K个保持不动； 如果剩下的不足K个，那么len个全部反转; */
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
using namespace std;
int main()
{
//这里测试用的，提交时写到 class {} 就行了
string s = "hyzqyljrnigxvdtneasepfahmtyhlohwxmkqcdfehybknvdmfrfvtbsovjbdhevlfxpdaovjgunjqlimjkfnqcqnajmebeddqsgl"; int k = 39;
char ch_temp;
if (s.length()<k) {
for (int i = 0; i<s.length() / 2; i++) {
ch_temp = s[i]; s[i] = s[s.length() - 1 - i]; s[s.length() - 1 - i] = ch_temp;
}
cout<<s;
}
else if (s.length() >= k && s.length()<2 * k) {
for (int i = 0; i<k / 2; i++) {
ch_temp = s[i]; s[i] = s[k - 1 - i]; s[k - 1 - i] = ch_temp;
}
cout<<s;
}
else {
int i = 0, t = 0;
while ( t * 2 * k <= s.length() - 1) {
if ((s.length() - (i + t * 2 * k)) >= k && (s.length() - (i + t * 2 * k))<2 * k) {
for (i = 0; i<k / 2; i++) {
ch_temp = s[i + t * 2 * k]; s[i + t * 2 * k] = s[t * 2 * k + k - 1 - i]; s[t * 2 * k + k - 1 - i] = ch_temp;
}
i = 0, t++;
}
else if ((s.length() - (i + t * 2 * k)) < k) {
int len = s.length() - (i + t * 2 * k);
for (i = 0; i<len / 2; i++) {
ch_temp = s[i + t * 2 * k]; s[i + t * 2 * k] = s[len-1 - i + t * 2 * k]; s[len - 1 - i + t * 2 * k] = ch_temp;
}
i = 0, t++;
}else {
for (i = 0; i < k / 2; i++) {
ch_temp = s[i + t * 2 * k]; s[i + t * 2 * k] = s[k - 1 - i + t * 2 * k]; s[k - 1 - i + t * 2 * k] = ch_temp;
}
i = 0, t++;
}
}
cout<<s;
}
return 0;
}