LeetCode之Reverse String II
【摘要】 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, ...
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:
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Input: s = "abcdefg", k = 2
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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
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public class Solution {
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public String reverse(String s) {
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if (s == null || s.length() == 0) {
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return null;
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}
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char[] chars = s.toCharArray();
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int length = chars.length;
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int start = 0, end = length - 1;
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while (start < end) {
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char tmp = chars[start];
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chars[start++] = chars[end];
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chars[end--] = tmp;
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}
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String result = "";
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for (char c : chars)
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result += c;
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return result;
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}
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public String reverseStr(String s, int k) {
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if (null == s || s.length() == 0)
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return null;
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int length = s.length();
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int one = 1;
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int count = length % k == 0 ? length / k : length / k + 1;
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String result = "";
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for (int i = 1; i <= count; i++) {
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if (one % 2 == 1)
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result += reverse(s.substring(k * (i -1), (k * i > length ? length : k * i)));
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else
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result += s.substring(k * (i -1), (k * i > length ? length : k * i));
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one++;
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}
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return result;
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}
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}
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
文章来源: chenyu.blog.csdn.net，作者：chen.yu，版权归原作者所有，如需转载，请联系作者。
原文链接：chenyu.blog.csdn.net/article/details/68238590
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