1、題目

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.

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Example:

Input: s = "abcdefg", k = 2 Output: "bacdfeg"

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Restrictions:

The string consists of lower English letters only.

Length of the given string and k will in the range [1, 10000]

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2、代碼實(shí)現(xiàn)

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;} }
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3、總結(jié)

先把問題化為一小步，分治的思想，比如我們先實(shí)現(xiàn)字符串的反轉(zhuǎn)功能，然后再更具條件來實(shí)現(xiàn)，哪些子字符串需要反轉(zhuǎn)

總結(jié)

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