【LeetCode】541. Reverse String II 解题报告（Python）

原创

负雪明烛 2022-02-10 15:09:14 博主文章分类：LeetCode ©著作权

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作者：负雪明烛

id： fuxuemingzhu

题目描述

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]

题目大意

每2k个字符的前k个字符进行翻转，然后后面k个数字正常，进行拼接到一起。

解题方法

Java解法

也是很简单的题，但是我竟然耽误了很久。主要问题出现在了reverse函数上。我犯了错误。因为这个i不是从0开始的，那么，end-1-i的时候一定要再加上start才可以，负责不是end的下一个字符。只要细心还是可以做对的，实在不行就得用debug了。

public class Solution {
    public String reverseStr(String s, int k) {
        char[] ans = s.toCharArray();
        int len = s.length();
        for (int i = 0; i < len; i += 2 * k) {
            if (len - i < k) {
                reverse(ans, i, len);
            } else {
                reverse(ans, i, i + k);
            }
        }
        return new String(ans);
    }
    public void reverse(char[] chars, int start, int end){
        for (int i = start; i < (start + end) / 2; i++) {
            char temp = chars[i];
            chars[i] = chars[end - 1 - i + start];
            chars[end - 1 - i + start] = temp;
        }
    }
}

Python解法

Python的切片就是做这个的！而且切片很友好，如果切到外边的话也无所谓，因为Python会把切到外边的自动过滤掉。

class Solution:
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        N = len(s)
        res = ""
        pos = 0
        while pos < N:
            nx = s[pos : pos + k]
            res = res + nx[::-1] + s[pos + k : pos + 2 * k]
            pos += 2 * k
        return res