leetcode-541. 反转字符串 II

一.题目描述

二.第1次提交(for循环，std::reverse)

class Solution {public:string reverseStr(string s, int k) {for (int i = 0; i < s.size(); i += 2 * k) {if ((s.size() - i) < k) {reverse(s.begin() + i, s.end());} else if ((s.size() - i) >= k && (s.size() - i) < 2 * k) {reverse(s.begin() + i, s.begin() + i + k);} else {reverse(s.begin() + i, s.begin() + i + k);}}return s;}
};

三.第2次提交

class Solution {public:string reverseStr(string s, int k) {for (int i = 0; i < s.size(); i += 2 * k) {if ((s.size() - (i + 1)) < k) {reverse(s.begin() + i, s.end());} else if ((s.size() - (i + 1)) < 2 * k && (s.size() - (i + 1)) >= k) {reverse(s.begin() + i, s.begin() + i + k);} else if ((i + 2 * k) < s.size()) {reverse(s.begin() + i, s.begin() + i + k);}}return s;}
};

四.第3次提交

class Solution {public:string reverseStr(string s, int k) {for (int i = 0; i < s.size(); i = i + 2 * k) {if (i + 2 * k < (s.size())) {reverse(s.begin() + i, s.begin() + i + k);} else if (s.size() - i < k) {reverse(s.begin() + i, s.end());} else {reverse(s.begin() + i, s.begin() + i + k);}}return s;}
};

五.第4次提交

class Solution {public:string reverseStr(string s, int k) {for (int i = 0; i < s.size(); i = i + 2 * k) {if (2 * k < (s.size() - i)) {reverse(s.begin() + i, s.begin() + i + k);} else if (s.size() - i < k) {reverse(s.begin() + i, s.end());} else {reverse(s.begin() + i, s.begin() + i + k);}}return s;}
};

六.代码随想录解答一

class Solution {
public:string reverseStr(string s, int k) {for (int i = 0; i < s.size(); i += (2 * k)) {// 1. 每隔 2k 个字符的前 k 个字符进行反转// 2. 剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符if (i + k <= s.size()) {reverse(s.begin() + i, s.begin() + i + k );} else {// 3. 剩余字符少于 k 个，则将剩余字符全部反转。reverse(s.begin() + i, s.end());}}return s;}
};

七.代码随想录解答二

class Solution {
public:void reverse(string& s, int start, int end) {for (int i = start, j = end; i < j; i++, j--) {swap(s[i], s[j]);}}string reverseStr(string s, int k) {for (int i = 0; i < s.size(); i += (2 * k)) {// 1. 每隔 2k 个字符的前 k 个字符进行反转// 2. 剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符if (i + k <= s.size()) {reverse(s, i, i + k - 1);continue;}// 3. 剩余字符少于 k 个，则将剩余字符全部反转。reverse(s, i, s.size() - 1);}return s;}
};

八.代码随想录解答三

class Solution {
public:string reverseStr(string s, int k) {int n = s.size(),pos = 0;while(pos < n){//剩余字符串大于等于k的情况if(pos + k < n) reverse(s.begin() + pos, s.begin() + pos + k);//剩余字符串不足k的情况 else reverse(s.begin() + pos,s.end());pos += 2 * k;}return s;}
};