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]
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 ""