[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]

反转字符串二。题意跟版本一很接近，多一个限制条件K。需要反转每2K个字符，如果剩下的字符长度不足，就整个反转他们。

这个题出的不是很好。题目不难但是题意讲的非常绕，差评很多。既然是反转每2K个字母，则正常逻辑会认为反转部分的轴是在第K个字母附近，但是根据例子，因为K = 2所以涉及到的字母是前四个ABCD，但是实际被反转的字母只有前K个，AB。所以这个题可以理解为每2K个字母，只反转前K个字母。

时间O(n^2)

空间O(n) - 因为convert成了数组

JavaScript实现

/**
 * @param {string} s
 * @param {number} k
 * @return {string}
 */
var reverseStr = function (s, k) {
    let t = s.split('');
    for (let i = 0; i < s.length; i += 2 * k) {
        let start = i;
        let end = i + k - 1;
        while (start < end) {
            const temp = t[start];
            t[start] = t[end];
            t[end] = temp;
            start++;
            end--;
        }
    }
    return t.join('');
};

 

Java实现

class Solution {
    public String reverseStr(String s, int k) {
        char[] input = s.toCharArray();
        for (int i = 0; i < input.length; i += 2*k) {
            int start = i;
            int end = Math.min(start + k - 1, input.length - 1);
            while (start < end) {
                char temp = input[start];
                input[start] = input[end];
                input[end] = temp;
                start++;
                end--;
            }
        }
        return String.valueOf(input);
    }
}

 

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