LeetCode之Reverse String II

1、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]

2、代码实现

public class Solution {
    public  String reverse(String s) {
if (s == null  || s.length() == 0) {
return null;
}
char[] chars = s.toCharArray();
int length = chars.length;
int start = 0, end = length - 1;
while (start < end) {
char tmp = chars[start];
 chars[start++] = chars[end];
 chars[end--] = tmp;
}
String result = "";
for (char c : chars)
result += c;
return result;
}

    public  String reverseStr(String s, int k) {
    if (null == s || s.length() == 0)
    return null;
    int length = s.length();
    int one = 1;
    int count = length % k == 0 ? length / k : length / k + 1;
    String result = "";
    for (int i = 1; i <= count; i++) {
    if (one % 2 == 1)
    result += reverse(s.substring(k * (i -1),  (k * i > length ? length : k * i)));
    else
    result += s.substring(k * (i -1),  (k * i > length ? length : k * i));
    one++;
    }
        return result;
    }
}

 

3、总结

先把问题化为一小步，分治的思想，比如我们先实现字符串的反转功能，然后再更具条件来实现，哪些子字符串需要反转