LeetCode题解之 Reverse Only Letters 1、题目描述 2、题目描述 利用栈实现逆序。 3、代码 1stringreverseOnlyLetters(stringS) {2if(S.size() ==0|| S.size() ==1)3returnS;45stack<string>st;6for(string::iterator it = S.begin(); it != S.end(); it++) {7if( isalpha(*it) ){8strings...
class Solution: def reverseOnlyLetters(self, S: str) -> str: letters = [c for c in S if c.isalpha()] res = [] for c in S: if c.isalpha(): res.append(letters.pop()) else: res.append(c) return "".join(res)
My in-place C++ solution: class Solution {public:string reverseOnlyLetters(string S) {int l = 0;int r = S.size();while (l < r) {if (!isalpha(S[l])) {l++;} else if (!isalpha(S[r-1])) {r--;} else {swap(S[l], S[r-1]);l++;r--;}}return S;}}; Read more 17 ...
Leetcode 917.仅仅反转字母(Reverse Only Letters) Leetcode 917.仅仅反转字母 1 题目描述(Leetcode题目链接) 给定一个字符串 S,返回 “反转后的” 字符串,其中不是字母的字符都保留在原地,而所有字母的位置发生反转。 提示: S.length <= 100 33 <= S[i].ASCIIcode <= 122 S 中不包含 \ or...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
Restrictions: The string consists of lower English letters only. 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 ...
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; } }; 参考资料: https://discuss.leetcode.com/topic/82652/one-line-c/2 ...
技术标签: 字符串 LeetCodeGiven 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...
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 public class Solution { public String reverse(String s) { if (s == null || s.length() == 0) { ...
leetcode.cn/problems/ev 解题方法 俺这版 class Solution { public int evalRPN(String[] tokens) { if (tokens == null) { return 0; } Stack<String> stack = new Stack<>(); for (int i = 0; i < tokens.length; i++) { if (tokens[i].equals("+")) { Integer item2 = Integer.val...