1publicclassSolution {2publicString reverseString(String s) {3if(s ==null|| s.length() <=1)returns;4StringBuffer res =newStringBuffer(s.length());5for(inti = s.length()-1;i >=0 ;i--){6res.append(s.charAt(i));7}8returnres.toString();9}10}...
java与C++遍历string和reverse用法的区别 概述Java遍历string 需要先调用toCharArray()转换成char数组,然后调用for循环。示例代码class Solution { public boolean isPalindrome(String s) { if(s.equals("")||s==null) return true; s = s.toLowerCase(); char[] ss = s.toCharArray(); int k = s....
}publicMain() {Scannerin=newScanner(System.in);while(in.hasNextLine()) {Stringstr =in.nextLine();Stringres =reverse(str);System.out.println(res); } }publicstaticvoidmain(String[] args) {Mainsolution =newMain(); } } AI代码助手复制代码...
1classSolution {2publicString reverseWords(String s) {3char[] s1 =s.toCharArray();4StringBuilder s2 =newStringBuilder();5intleft = 0, right = s.length()-1;6//去掉字符串左右两端的空白7while(s1[left] == ' ') {8left++;9}10while(s1[right] == ' ') {11right--;12}13//添加单词...
注意string与int间的转换,string[] - '0' 变成int,int + '0' 变成 string[]。 由于我们希望能用下标与数字的位数对应起来,所以需要用 reverse(s.begin(), s.end()) 对String翻转一下存储,最后翻转回来,并将得到的结果的无用高位 '0' 删除掉,如果结果为 '0',还要特殊处理。
public class Solution { public String reverseWords(String s) { // 使用trim()方法去除字符串前后的空格 s = s.trim(); // 使用StringBuilder来存储结果,避免频繁的字符串拼接操作 StringBuilder sb = new StringBuilder(); // 定义一个双指针,分别指向字符串的开头和末尾 int start = 0; int end = s...
选几种典型的加以说明 import java.io.BufferedInputStream; import java.util.Scanner; public class test { //...s.substring(N / 2, N); return reverseString1(s2) + reverseString1(s1); } // 返回串,系统自带方法就是这样实现的...long l2 = System.nanoTime(); System.out.println(reverseString...
publicclassCrunchifyReverseString{ publicstaticvoidmain(String[]args){ StringtestString ="Crunchify.com Example"; System.out.println("String: "+ testString); System.out.println("\nSolution1: Reverse Using reverseStringBuffer: "+reverseStringBuffer(testString)); ...
class Solution { public String reverseWords(String s) { if (s == null || s.length() == 0) { return ""; } // split to words by space String[] arr = s.split(" "); StringBuilder sb = new StringBuilder(); for (int i = arr.length - 1; i >= 0; --i) { if (!arr[i...
class Solution { public boolean isPalindrome(int x) { String s = String.valueOf(x); StringBuilder sb = new StringBuilder(s); sb.reverse(); return sb.toString().equals(s); } } 1. 2. 3. 4. 5. 6. 7. 8. 时间复杂度:数字的位数,数字大约有位,翻转操作要执行循环。复杂度为 ...