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:
- The string consists of lower English letters only.
- 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
|
