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]
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
1 /** 2 * @param {string} s 3 * @param {number} k 4 * @return {string} 5 */ 6 var reverseStr = function (s, k) { 7 let t = s.split(''); 8 for (let i = 0; i < s.length; i += 2 * k) { 9 let start = i; 10 let end = i + k - 1; 11 while (start < end) { 12 const temp = t[start]; 13 t[start] = t[end]; 14 t[end] = temp; 15 start++; 16 end--; 17 } 18 } 19 return t.join(''); 20 };
Java实现
1 class Solution { 2 public String reverseStr(String s, int k) { 3 char[] input = s.toCharArray(); 4 for (int i = 0; i < input.length; i += 2*k) { 5 int start = i; 6 int end = Math.min(start + k - 1, input.length - 1); 7 while (start < end) { 8 char temp = input[start]; 9 input[start] = input[end]; 10 input[end] = temp; 11 start++; 12 end--; 13 } 14 } 15 return String.valueOf(input); 16 } 17 }
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