541. Reverse String II（LeetCode）

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:

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

   class Solution {
   public:
       string reverseStr(string s, int k) {
               int len = s.size();
           string s1 = "";
           if (len<k)
           {
               for (int i = len - 1; i >= 0; i--)
               {
                   s1 += s[i];
               }
           }
           else
           {
               for (int i = 0; i < len; i = i + 2*k)
               {
                 for(int j=(i+k-1>len?len-1:(i+k-1));j>=i;j--)
                 {
                     s1+=s[j];
                 }
                 for(int j=i+k;j<((i+2*k)>len?len:(i+2*k));j++)
                 {
                     s1+=s[j];
                 }
               }
           }
           return s1;
       }
   };

   我的代码普遍有个特点，就是暴力，哎，什么时候才能学会取巧呢？下面的是比较好的算法。

   class Solution {
       void reverse(string &s, int l, int r)
       {
           if (r > (s.length() - 1))
               r = s.length() - 1;
           while (l < r)
           {
               swap(s[l], s[r]);
               l++;
               r--;
           }
       }
   public:
       string reverseStr(string s, int k) {
   
           for (int i = 0; i<s.length(); i += 2 * k)
           {
               reverse(s, i, i + k - 1);
           }
   
           return s;
       }
   };