problem 541. Reverse String II 题意: 给定一个字符串,每隔k个字符翻转这k个字符,剩余的小于k个则全部翻转,否则还是只翻转剩余的前k个字符。 solution1: classSolution {public:stringreverseStr(strings,intk) {intn =s.size();intcnt = n /k;for(inti=0; i<=cnt; i++) {if(i%2==0) {if(i...
classSolution {public:stringreverseStr(strings,intk) {for(inti =0; i < s.size(); i +=2*k) { reverse(s.begin()+ i, min(s.begin() + i +k, s.end())); }returns; } }; 类似题目: Reverse String 参考资料: https://discuss.leetcode.com/topic/82652/one-line-c/2 https://disc...
Reverse String II 自己的解法: 就是用一个StringBuffer来进行字符串的接收,利用一个指针来指示当前是哪一个k部分,当 i + k < s.length() 时,证明从 i 到 i + k - 1部分是可以进行反转的,而从i + k 部分到 i + 2 * k部分是直接进行拼接,利用条件 j < s.length()即可进行限制,不过对于最后剩...
Leetcode 541. Reverse String II 原题链接:https://leetcode.com/problems/reverse-string-ii/description/ 描述: 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......
1classSolution {2publicString reverseStr(String s,intk) {3char[] array =s.toCharArray();4for(inti = 0 ; i < s.length() ; i += 2*k ){5reverse(array,i,i+k-1);6}7returnString.valueOf(array);8}910privatevoidreverse(char[] array,intstart,intend) {11if( start > array.length...
输入: s ="abcdefg", k = 2 输出:"bacdfeg" 要求: 该字符串只包含小写的英文字母。 给定字符串的长度和 k 在[1, 10000]范围内。 解法: classSolution{public:// method 1:stringreverseStr1(string s,intk){intsz = s.size();inti =0, j =0;boolflip =true; ...
LeetCode:Reverse String II 541. Reverse String II 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 ...
1classSolution {2publicString reverseStr(String s,intk) {3char[] input =s.toCharArray();4for(inti = 0; i < input.length; i += 2*k) {5intstart =i;6intend = Math.min(start + k - 1, input.length - 1);7while(start <end) {8chartemp =input[start];9input[start] =input[end...
Length of the given string and k will in the range [1, 10000] 本题题意十分简单,就是做一个字符串的反转,注意是没2k个子串中前k个反转,其余的不变 建议和leetcode 344. Reverse String 反转字符串 一起学习 代码如下: #include <iostream> ...
利用Reverse String 里的 two pointers 来反转k 个chars,每一次增加 i by 2*k 就可以了。 具体请看code。 Java Solution: Runtime beats 67.31% 完成日期:04/07/2018 关键词:two pointers 关键点:swap left and right chars 1classSolution2{3publicString reverseStr(String s,intk)4{5char[] c_arr =...