翻转链表查找链表的倒数第k个节点位运算加法

反转单链表

void Reverse(ListNode *&pHead)
{
    if (pHead == nullptr||pHead->_pNext==nullptr)
        return;
    ListNode *pCur = pHead;
    ListNode *pNewHead = nullptr;
    ListNode *pTemp = nullptr;
    while

查找单链表的倒数第k个节点，要求只能遍历一次链表

ListNode *FindKNode(ListNode *&pHead,int k)
{
    if (pHead == nullptr)
        return nullptr;
    ListNode *pCur = pHead;
    ListNode *pFast = pHead;
    ListNode *pSlow = pHead;
    for (int i = 0; i < k-1; i++)
    {
        if (pFast->_pNext != nullptr)
            pFast = pFast->_pNext;
        else
            return nullptr;
    }
    while (pFast->_pNext != nullptr)
    {
        pFast = pFast->_pNext;
        pSlow = pSlow->_pNext;
    }
    return

加法器的实现

递归法：
#include<iostream>
using namespace std;
int getNum(int a, int b)
{
    int wei = a ^ b;
    int jinwei = (a&b) <<1;
    if (jinwei == 0)
        return wei;
    else
        return getNum(wei, jinwei);
}
int main()
{
    int a;
    int b;
    while (cin>>a&&cin>>b)
        cout

#include <iostream>
using namespace std;
int add1(int x, int y)
{
    int sum;
    int carry;
    int bx, by;
    int base;
    base = 1;
    //carry用来判断是否需要进位
    carry = 0;
    sum = 0;
    while (base != 0) {
        bx = x & base;
        by = y & base;
        base <<= 1;
        sum |= ((bx) ^ (by) ^ carry);
        carry = ((bx & by) || (bx & carry) || (by & carry)) ? base : 0;
    }
    return sum;
}