LeetCode 541. Reverse String II（C++版）

题目描述：

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from
the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

Example:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Restrictions:
The string consists of lower English letters only.
Length of the given string and k will in the range [1, 10000]

思路分析：

把s按照2k长度为单位处理，最后再处理一个不足2k长度的。每一个2k长度中，得到两个k长度的子串，前k个反转，后k个不变，加入结果字符串s中。

代码：

class Solution {
public:
string reverseWord(string &s) {

int len = s.length();
for(int i = 0; i < len/2; i ++) {
char tmp = s[i];
s[i] = s[len-i-1];
s[len-1-i] = tmp;
}
return s;
}
string reverseStr(string s, int k) {
if(k <= 1) return s;

string res="";
int len = s.length();
int p = 2*k; //2k个字符为一组，每组中前k个需反转
int cnt = len / p; //一共有cnt个2k长度的字符串
int remaider = len % p; //最后剩下的一组，不足2k个字符

//先处理前cnt个2k长度的字符串
for(int i = 0; i < cnt; i ++) {
string s1 = s.substr(p*i, k);
string s2 = s.substr(p*i+k, k);
reverseWord(s1);
res += s1 + s2;
}
//处理剩下的一个不足2k的字符串
if(remaider <= k){
string s1 = s.substr(cnt*p);
reverseWord(s1);
res += s1;
}
else {
string s1 = s.substr(cnt*p, k);
string s2 = s.substr(cnt*p+k);
reverseWord(s1);
res += s1 + s2;
}
return res;
}
};
Runtime: 9
ms