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Merge pull request #276 from zqh1059405318/master

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更新0541.反转字符串II Java版本
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Carl Sun
2021-05-28 20:58:46 +08:00
committed by GitHub
parent 847a475aba 7f3b661add
commit c23b7b6804
1 changed files with 15 additions and 18 deletions
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33
problems/0541.反转字符串II.md
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@ -106,27 +106,24 @@ Java：
class Solution {
public String reverseStr(String s, int k) {
StringBuffer res = new StringBuffer();
for (int i = 0; i < s.length(); i += (2 * k)) {
int length = s.length();
int start = 0;
while (start < length) {
// 找到k处和2k处
StringBuffer temp = new StringBuffer();
// 剩余字符大于 k 个，每隔 2k 个字符的前 k 个字符进行反转
if (i + k <= s.length()) {
// 反转前 k 个字符
temp.append(s.substring(i, i + k));
res.append(temp.reverse());
// 与length进行判断，如果大于length了，那就将其置为length
int firstK = (start + k > length) ? length : start + k;
int secondK = (start + (2 * k) > length) ? length : start + (2 * k);
// 反转完前 k 个字符之后，如果紧接着还有 k 个字符，则直接加入这 k 个字符
if (i + 2 * k <= s.length()) {
res.append(s.substring(i + k, i + 2 * k));
// 不足 k 个字符，则直接加入剩下所有字符
} else {
res.append(s.substring(i + k, s.length()));
}
continue;
}
// 剩余字符少于 k 个，则将剩余字符全部反转。
temp.append(s.substring(i, s.length()));
//无论start所处位置，至少会反转一次
temp.append(s.substring(start, firstK));
res.append(temp.reverse());
// 如果firstK到secondK之间有元素，这些元素直接放入res里即可。
if (firstK < secondK) { //此时剩余长度一定大于k。
res.append(s.substring(firstK, secondK));
}
start += (2 * k);
}
return res.toString();
}