LeetCode //C - 541. Reverse String II

541. Reverse String II

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 . l e n g t h < = 1 0 4 1 <= s.length <= 10^4 1<=s.length<=104
s consists of only lowercase English letters.
1 < = k < = 1 0 4 1 <= k <= 10^4 1<=k<=104

From: LeetCode

Link: 541. Reverse String II

Solution:

Ideas:

1. Helper Function (reverse):

This function reverses the characters in the substring of s from index start to end.

2. Main Function (reverseStr):

It iterates through the string in segments of 2k.
For every 2k segment, the first k characters are reversed. The rest remain unchanged.
If there are fewer than k characters left, reverse all of them.
If there are between k and 2k characters, reverse the first k and leave the rest unchanged.

3. Edge Cases:

When k is greater than the remaining length of the string, it handles it by only reversing up to the string's end.
The function is efficient and adheres to the constraints, as the operations are performed in linear time relative to the string length.

Code:

c 复制代码

void reverse(char* s, int start, int end) {
    while (start < end) {
        char temp = s[start];
        s[start] = s[end];
        s[end] = temp;
        start++;
        end--;
    }
}

char* reverseStr(char* s, int k) {
    int len = strlen(s);
    for (int i = 0; i < len; i += 2 * k) {
        // Reverse the first k characters in the current segment
        int end = (i + k - 1 < len) ? i + k - 1 : len - 1;
        reverse(s, i, end);
    }
    return s;
}