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合并排序链表
不改变原链表
#include<iostream>
using namespace std;
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :val(x), next(NULL) {}
};
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1==NULL)return pHead2;
if(pHead2==NULL)return pHead1;
ListNode *r, *p1=pHead1, *p2 = pHead2, *tail;
if(p1->val>p2->val){
r=new ListNode(p2->val);
tail = r;
p2=p2->next;
}
else{
r=new ListNode(p1->val);
tail = r;
p1=p1->next;
}
while((p1!=NULL)&&(p2!=NULL)){
if(p1->val>p2->val){
ListNode* tmp = new ListNode(p2->val);
tail->next = tmp;
p2 = p2->next;
}
else{
ListNode* tmp = new ListNode(p1->val);
tail->next = tmp;
p1 = p1->next;
}
tail = tail->next;
}
while(p1!=NULL){
ListNode* tmp = new ListNode(p1->val);
tail->next = tmp;
tail = tail->next;
p1 = p1->next;
}
while(p2!=NULL){
ListNode* tmp = new ListNode(p2->val);
tail->next = tmp;
tail = tail->next;
p2 = p2->next;
}
return r;
}
};
int main()
{
Solution x;
ListNode *l1=new ListNode(1);
ListNode *l2=new ListNode(2);
ListNode *l3=new ListNode(3);
ListNode *l4=new ListNode(4);
ListNode *l5=new ListNode(5);
ListNode *l6=new ListNode(6);
l5->next=l6;l4->next=l5;l3->next=l4;l2->next=l3;l1->next=l2;
ListNode *r1=new ListNode(1);
ListNode *r2=new ListNode(2);
ListNode *r3=new ListNode(3);
ListNode *r4=new ListNode(4);
ListNode *r5=new ListNode(5);
ListNode *r6=new ListNode(6);
r5->next=l6;r4->next=r5;r3->next=r4;r2->next=r3;r1->next=r2;
while(l2!=NULL){
cout<<l2->val;
l2=l2->next;
}cout<<endl;
while(r2!=NULL){
cout<<r2->val;
r2=r2->next;
}cout<<endl;
ListNode *r = x.Merge(l1,r1);
ListNode *display=r;
while(display!=NULL){
cout<<display->val;
display=display->next;
}cout<<endl;
return 0;
}
直接操作原链表
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1==NULL)return pHead2;
if(pHead2==NULL)return pHead1;
ListNode *r, *tail;
if(pHead1->val>pHead2->val){
r=pHead2;
pHead2=pHead2->next;
}
else{
r=pHead1;
pHead1=pHead1->next;
}
tail=r;
while((pHead1!=NULL)&&(pHead2!=NULL)){
if(pHead1->val>pHead2->val){
tail->next=pHead2;
pHead2=pHead2->next;
}
else{
tail->next=pHead1;
pHead1=pHead1->next;
}
tail=tail->next;
}
if(pHead1!=NULL)tail->next=pHead1;
if(pHead2!=NULL)tail->next=pHead2;
return r;
}
};
链表中倒数第k个节点
要点：双指针
class Solution {
public:
ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
if(pListHead==NULL||k==0)return NULL;
ListNode *pl=pListHead, *pr=pListHead;
for(int i=0;i<k;i++)pr=pr->next;
while(pr!=NULL){
pr=pr->next;
pl=pl->next;
}
return pl;
}
};
反转链表
class Solution {
public:
ListNode* ReverseList(ListNode* pHead) {
if(pHead==NULL)return NULL;
ListNode *r=NULL, *tmp;
while(pHead!=NULL){
tmp=pHead;
pHead=pHead->next;
tmp->next=r;
r=tmp;
}
return r;
}
};
数值的整数次方
要点：考虑负数指数
class Solution {
public:
double Power(double base, int exponent) {
if(exponent==0)return 1;
int pos;
if(exponent>0)pos=1;
else {
pos=-1;
exponent=-exponent;
}
double r=1;
for(int i=0;i<exponent;i++)r*=base;
if(pos==1)return r;
else return 1/r;
}
};
包含min函数的栈
要点：两个栈，同入同出
class Solution {
private:
stack<int> s1,s2;
public:
void push(int value) {
s1.push(value);
if(s2.empty())s2.push(value);
else if(s2.top()<value)s2.push(s2.top());
else s2.push(value);
}
void pop() {
s1.pop();
s2.pop();
}
int top() {
return s1.top();
}
int min() {
return s2.top();
}
};
两个转轮密码锁，求最少转动次数使匹配
#include <iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
int n;
vector<int> n1, n2;
cin >> n;
string input;
cin >> input;
for (int i = 0;i < n;i++) {
n1.push_back(input[i] - '0');
}
cin >> input;
for (int i = 0;i < n;i++) {
n2.push_back(input[i] - '0');
}
int r = 0;
for (int i = 0;i < n;i++) {
int once = n1[i] - n2[i];
if (once < 0)once = - once;
if (once > 5)once = 10 - once;
r += once;
}
cout << r;
return 0;
}
a,b二人各自有两个等长数组，二人轮流执行一下操作之一：从对方的数组中删去1个数；从自己的数组中删去一个数，并累加到自己的得分。求两个数组删完时a比b最多多多少分（a先操作）
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
bool comp(const int &a, const int &b)
{
return a > b;
}
int main()
{
int n;
cin >> n;
vector<int>n1, n2;
int p;
for (int i = 0;i < n;i++) {
cin >> p;
n1.push_back(p);
}
for (int i = 0;i < n;i++) {
cin >> p;
n2.push_back(p);
}
sort(n1.begin(), n1.end(), comp);
sort(n2.begin(), n2.end(), comp);
cout << endl;
for (int i = 0;i < n;i++)cout << n1[i];
for (int i = 0;i < n;i++)cout << n2[i];
cout << endl;
int p1 = 0, p2 = 0;
int i1 = 0, i2 = 0;
int turn = 0;
while ((i1 < n) && (i2 < n)) {
if (turn == 0) {
if (n1[i1] == n2[i2]) {
i1++;
i2++;
continue;
}
else if (n1[i1] > n2[i2]) {
turn = 1;
p1 += n1[i1];
i1++;
}
else if (n1[i1] < n2[i2]) {
turn = 1;
i2++;
}
}
else {
if (n1[i1] == n2[i2]) {
i1++;
i2++;
continue;
}
else if (n1[i1] > n2[i2]) {
turn = 0;
i1++;
}
else if (n1[i1] < n2[i2]) {
turn = 0;
p2 += n2[i2];
i2++;
}
}
}
while (i1 < n) {
if (turn == 0) {
p1 += n1[i1];
i1 += 2;
}
else {
i1++;
turn = 0;
}
}
while (i2 < n) {
if (turn == 1) {
p2 += n2[i2];
i2 += 2;
}
else {
i2++;
turn = 1;
}
}
cout << p1 - p2;
return 0;
}