题目地址：https://leetcode.com/problems/reverse-string-ii/description/

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]

 

给定一个字符串和一个整数 k，你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转。如果剩余少于 k 个字符，则将剩余的所有全部反转。如果有小于 2k 但大于或等于 k 个字符，则反转前 k 个字符，并将剩余的字符保持原样。

示例:

输入: s = "abcdefg", k = 2
输出: "bacdfeg"

要求:

  1. 该字符串只包含小写的英文字母。
  2. 给定字符串的长度和 k 在[1, 10000]范围内。

 

 

一种比较好的写法(当然换成字符数组交换顺序就更好了)

class Solution {
    public String reverseStr(String s, int k) {
        StringBuilder str = new StringBuilder();
        int len = s.length();
        int num = 0;
        for (int i = 0; i < len; i += k) {
            if ((num & 1) == 0) {
                str.append(new StringBuilder(s.substring(i, Math.min(i + k, len))).reverse());
            } else {
                str.append(s.substring(i, Math.min(i + k, len)));
            }
            ++num;
        }
        return new String(str);
    }
}

 

 

Debug code in playground：

class Solution {
    public String reverseStr(String s, int k) {
        StringBuilder str = new StringBuilder();
        int len = s.length();
        int num = 0;
        for (int i = 0; i < len; i += k) {
            if ((num & 1) == 0) {
                str.append(new StringBuilder(s.substring(i, Math.min(i + k, len))).reverse());
            } else {
                str.append(s.substring(i, Math.min(i + k, len)));
            }
            ++num;
        }
        return new String(str);
    }
}

public class MainClass {
    public static String stringToString(String input) {
        if (input == null) {
            return "null";
        }
        return Json.value(input).toString();
    }
    
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String line;
        while ((line = in.readLine()) != null) {
            String s = stringToString(line);
            line = in.readLine();
            int k = Integer.parseInt(line);
            
            String ret = new Solution().reverseStr(s, k);
            
            String out = (ret);
            
            System.out.print(out);
        }
    }
}

以下是最初自己想到的糟糕算法，勉强能解决

class Solution {
    public String reverseStr(String s, int k) {
        StringBuilder s1 = new StringBuilder(s);
        int n = k;
        StringBuilder ss = new StringBuilder();
        if (n == 1) {
            return s;
        } else {
            if (s.length() < n) {
                return new String(s1.reverse());
            } else {
                int t1 = 0, t2 = 0, num = n - 1, temp = 0;
                boolean flag = true;
                int len = s.length();
                for (int i = 0; i < len; ++i) {
                    if (t1 <= num) {
                        ++t1;
                        t2 = 0;
                        if (flag) {
                            temp = i + num;
                            flag = false;
                        }
                        if (temp < len) { // 剩下的足够k
                            ss.append(s1.charAt(temp--));
                        } else { // 剩下的不足k
                            for (int j = len - 1; j >= i; --j) {
                                ss.append(s1.charAt(j));
                            }
                            break;
                        }
                    } else if (t2 <= num) {
                        flag = true;
                        ++t2;
                        ss.append(s1.charAt(i));
                    }
                    if (t2 == n) {
                        t1 = 0;
                    }
                }
            }
            return new String(ss);
        }
    }
}

 

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