Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /.Each operand may be an integer or another expression. Some examples: ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 之前已经讲过逆波兰表达式的生成和运算方法，实际上计算是更为简单的。具体见博客。 package com.liuhao.acm.leetcode;import java.util...
转载请注明出处：z_zhaojun的博客 原文地址：http://blog.csdn.net/u012975705/article/details/50408975 题目地址：https://leetcode.com/problems/reverse-linked-list-ii/Reverse Linked List IIReverse a linked list from position m to n.Do it in-placeandin one-pass.For example: Given 1->2->3->4->5->NULL, m =2and n =4,return1->4->3->2->5->NULL.Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ ...
Problem: Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nodes itself may be changed. Only constant memory is allowed. For example, Given this linked list: 1->2->3->4->5 For k = 2, you should return: 2...
1publicstatic ListNode reverseBetween(ListNode head, int m, int n) {2 ListNode pre=head,current=head,mPre = new ListNode(0),mNode = new ListNode(0),nNode = new ListNode(0),temp;3 mPre.next=head;4int i=1;5while(i<=n)6 {7if(i==m-1)8 mPre=current;9if(i==m) 10 mNode=current; 11if(i==n) 12 nNode=current; 13if(m<i&&i<=n) 14 ...
Evaluate the value of an arithmetic expression in Reverse Polish Notation.Valid operators are +, -, *, /. Each operand may be an integer or another expression.Some examples: ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 就是求逆波兰表达式（后续遍历）的结果。1、直接求解，很慢publicclass Solution {publicint evalRPN(String[] tokens) {int len = tok...
【151-Evaluate Reverse Polish Notation（计算逆波兰式）】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】原题 Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Some examples: ["2", "1", "+", "3", "*"] ->((2 + 1) * 3) ->9["4", "13", "5", "/", "+"] ->(4 + (13 / 5)) ->6题目大意 ...
Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh". Subscribe to see which companies asked this question 1publicclass Solution {2public String reverseString(String s) {3if(s == null || s.length() <=1) return s;4 StringBuffer res = new StringBuffer(s.length());5for(int i = s.length()-1;i >=0 ;i--){6 res.appen...
题目链接:https://buuoj.cn/challenges#Java%E9%80%86%E5%90%91%E8%A7%A3%E5%AF%86 题目是个.class文件，拖进jd-gui里,看到java源码是个简单的加密，写个脚本跑一下 key = [ 180, 136, 137, 147, 191, 137, 147, 191, 148, 136, 133, 191, 134, 140, 129, 135, 191, 65 ] flag = "" for i in key:flag += chr(i - ord("@") ^ 0x20) print(flag)flag:This_is_the_flag_!
【LeetCode】150. Evaluate Reverse Polish Notation 逆波兰表达式求值（Medium）（JAVA） 题目地址: https://leetcode.com/problems/evaluate-reverse-polish-notation/ 题目描述： Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note: Division between two integers should truncate toward zero. The ...
1. Description:2.Solutions: 1 /**2 * Created by sheepcore on 2019-02-243 */4 class Solution {5 public String reverseWords(String s) {6 String[] splitStr = s.split(" "); 7 String temp = "";8 for (String str : splitStr) 9 temp += new StringBuffer(str).reverse().toString() + " "; 10 return temp.substring(0, temp.length() - 1); 11 } 12 }
题目描述： 反转一个单链表。 示例: 输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL 迭代解法 /** Definition for singly-linked list. public class ListNode {int val;ListNode next;ListNode(int x) { val = x; } }*/ class Solution {public ListNode reverseList(ListNode head) {ListNode pre = null;ListNode next = null;while(head!=null){next = head.next;head.next = pre;pre = head;head = next;}return pre;}...
Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/ class Solution {public ListNode reverseList(ListNode head) {ListNode curr = head;ListNode preNode = null;//反转后前节点ListNode nextTemp = null; //next nodewhile (curr != null){nextTemp = curr.next;/...
我对以下问题感兴趣：Collections.reverse()vs Lists.reverse()哪个更快？ 谢谢.解决方法:他们做不同的事情. Collections.reverse采用可变列表并反转其顺序.它需要线性时间.它必须. Guava的Lists.reverse返回一个反转列表的视图.它以恒定时间返回,但您将为每个操作支付视图的(小)开销.
Leetcode 344：Reverse String 反转字符串 公众号：爱写bug Write a function that reverses a string. The input string is given as an array of characters char[]. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. You may assume all the characters consist of printable ascii characters 编写一个函数，其作用是将输入的字符串反转过来。...
public int reverse(int x) {long res = 0;while (x != 0){res = res* 10 + x % 10;x /= 10;}if(res >= Integer.MAX_VALUE || res < Integer.MIN_VALUE)return 0;return (int)res;}