LeetCode 541. Reverse String II（反转字符串 II）

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:

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

解法一

思路

这题挺简单的，就是每隔k个字符就翻转k个字符，框架就是：

for (int i = 0; i < length; i += 2 * k) {
    if (i + k >= length) {
        reverse(str, i, length - 1);
    } else {
        reverse(str, i, i + k - 1);
    }
}

而reverse(str, start, end)就是简单的反转字符串了。

Python

class Solution(object):
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        length = len(s)
        for i in range(0, length, 2 * k):
            if i + k >= length:
                s = s[:i] + s[i:][::-1]
            else:
                s = s[:i] + s[i:i + k][::-1] + s[i + k:]
        return s

Java

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

C++

class Solution {
public:
    string reverseStr(string s, int k) {
        class Utils {
        public:
            void reverse(string &str, int start, int end) {
                while (start < end) {
                    char temp = str[start];
                    str[start] = str[end];
                    str[end] = temp;
                    start++;
                    end--;
                }
            }
        };
        Utils utils;
        int length = s.length();
        for (int i = 0; i < length; i += 2 * k) {
            if (i + k >= length) {
                utils.reverse(s, i, length - 1);
            } else {
                utils.reverse(s, i, i + k - 1);
            }
        }
        return s;
    }
};