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:

1. The string consists of lower English letters only.
2. 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

public class Solution {
    public String reverseStr(String s, int k) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length() / k + 1; i++) {
            if (i % 2 == 0) {
                StringBuilder st = new StringBuilder(s.substring(i * k, Math.min((i + 1) * k, s.length())));
                sb.append(st.reverse());
            } else {
                StringBuilder st = new StringBuilder(s.substring(i * k, Math.min((i + 1) * k, s.length())));
                sb.append(st);
            }
        }
        return sb.toString();
    }
}

别人的算法，只用char[]不用stringbuilder

https://discuss.leetcode.com/topic/82626/java-concise-solution

public class Solution {
    public String reverseStr(String s, int k) {
        char[] arr = s.toCharArray();
        int n = arr.length;
        int i = 0;
        while(i < n) {
            int j = Math.min(i + k - 1, n - 1);
            swap(arr, i, j);
            i += 2 * k;
        }
        return String.valueOf(arr);
    }
    private void swap(char[] arr, int l, int r) {
        while (l < r) {
            char temp = arr[l];
            arr[l++] = arr[r];
            arr[r--] = temp;
        }
    }
}

更多讨论：

https://discuss.leetcode.com/category/693/reverse-string-ii