leetcode 541. Reverse String II

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(object):
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        ans = []
        for i in xrange(0, len(s), k*2):
            ans.append(s[i:i+k][::-1])
            ans.append(s[i+k:i+2*k])            
        return "".join(ans)

class Solution(object):
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        s = list(s)
        for i in xrange(0, len(s), 2*k):
            s[i:i+k] = reversed(s[i:i+k])
        return "".join(s)

如果是java则，

public class Solution {
    public String reverseStr(String s, int k) {
        char[] arr = s.toCharArray();
        int n = arr.length;
        int i = 0;
        while(i < n) {
            int j = Math.min(i + k - 1, n - 1);
            swap(arr, i, j);
            i += 2 * k;
        }
        return String.valueOf(arr);
    }
    private void swap(char[] arr, int l, int r) {
        while (l < r) {
            char temp = arr[l];
            arr[l++] = arr[r];
            arr[r--] = temp;
        }
    }
}

还可以写成递归：

class Solution(object):
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        return s[:k][::-1]+s[k:2*k]+self.reverseStr(s[2*k:],k) if s else ""