541. Reverse String II

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

Python Code
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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)

Java Code
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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);
    }
}

C++ Code
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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;
    }
};

Go Code
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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)
}