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541. Reverse String II
题目 :
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]
链接:https://leetcode.com/problems/reverse-string-ii/#/description
4/1/2017
30ms
注意:
1. String如何转为StringBuilder,第6,9行
2. 第6,9行最后要跟s.length()判断而不是s.length()-1
1 public class Solution { 2 public String reverseStr(String s, int k) { 3 StringBuilder sb = new StringBuilder(); 4 for (int i = 0; i < s.length() / k + 1; i++) { 5 if (i % 2 == 0) { 6 StringBuilder st = new StringBuilder(s.substring(i * k, Math.min((i + 1) * k, s.length()))); 7 sb.append(st.reverse()); 8 } else { 9 StringBuilder st = new StringBuilder(s.substring(i * k, Math.min((i + 1) * k, s.length()))); 10 sb.append(st); 11 } 12 } 13 return sb.toString(); 14 } 15 }
别人的算法,只用char[]不用stringbuilder
https://discuss.leetcode.com/topic/82626/java-concise-solution
1 public class Solution { 2 public String reverseStr(String s, int k) { 3 char[] arr = s.toCharArray(); 4 int n = arr.length; 5 int i = 0; 6 while(i < n) { 7 int j = Math.min(i + k - 1, n - 1); 8 swap(arr, i, j); 9 i += 2 * k; 10 } 11 return String.valueOf(arr); 12 } 13 private void swap(char[] arr, int l, int r) { 14 while (l < r) { 15 char temp = arr[l]; 16 arr[l++] = arr[r]; 17 arr[r--] = temp; 18 } 19 } 20 }
更多讨论:
https://discuss.leetcode.com/category/693/reverse-string-ii
541. Reverse String II
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