Question

541. Reverse String II

中文文档

Description

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

If there are fewer 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 leave the other as original.

 

Example 1:

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

Example 2:

Input: s = "abcd", k = 2
Output: "bacd"

 

Constraints:

• 1 <= s.length <= 104
• s consists of only lowercase English letters.
• 1 <= k <= 104

Solutions

Solution 1

class Solution:
  def reverseStr(self, s: str, k: int) -> str:
    t = list(s)
    for i in range(0, len(t), k << 1):
      t[i : i + k] = reversed(t[i : i + k])
    return ''.join(t)

class Solution {
  public String reverseStr(String s, int k) {
    char[] chars = s.toCharArray();
    for (int i = 0; i < chars.length; i += (k << 1)) {
      for (int st = i, ed = Math.min(chars.length - 1, i + k - 1); st < ed; ++st, --ed) {
        char t = chars[st];
        chars[st] = chars[ed];
        chars[ed] = t;
      }
    }
    return new String(chars);
  }
}

class Solution {
public:
  string reverseStr(string s, int k) {
    for (int i = 0, n = s.size(); i < n; i += (k << 1)) {
      reverse(s.begin() + i, s.begin() + min(i + k, n));
    }
    return s;
  }
};

func reverseStr(s string, k int) string {
  t := []byte(s)
  for i := 0; i < len(t); i += (k << 1) {
    for st, ed := i, min(i+k-1, len(t)-1); st < ed; st, ed = st+1, ed-1 {
      t[st], t[ed] = t[ed], t[st]
    }
  }
  return string(t)
}