21. Merge Two Sorted Lists
Easy
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Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4
Although this is a easy question,I'm not familiar with nodelist operator.
This code clear and nest.Perfect thing is that it add a node in front of the List,So many NULL
situation we needn't consider.So in the end we only need to return first->next,It's what we need
But this use to new ListNode each time it need toevaluation,It's not solve the input problem by
the root.As I have mentioned before,it does not embody the idea of programming from special to
general!
#include "test1.h" #include <iostream> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution{ public: //LeetCode中的方法名称 ListNode* mergeTwoLists(ListNode* l1,ListNode* l2) { ListNode* head = new ListNode(0);//head指针是每次后移的 ListNode* firstNode = head;//需要确定好一个不变的头结点，然后返回这个头结点的next地址，才能返回完成的链表 while(l1!=NULL && l2!=NULL){ if(l1->val > l2->val){ head->next = l2;//这里只是确定好head->next是哪个结点，head还没有指向这个结点 l2 = l2->next; }else{ head->next = l1; l1 = l1->next; } head = head ->next;//head指向这个结点 }//while head->next = l1?l1:l2;//遇到一个为空后，哪个不为空就next指向哪一个 //return head->next;//输出是：7 8 ，因为head指针也在往后移动 return firstNode->next; } }; ListNode* initList1(ListNode* head){//初始化第一个链表 ListNode* node1 = new ListNode(2); head->next = node1; ListNode* node2 = new ListNode(3); node1->next = node2; ListNode* node3 = new ListNode(6); node2->next = node3; ListNode* node4 = new ListNode(8); node3->next = node4; return head; }//initList1 ListNode* initList2(ListNode* head){//初始化第二个链表 ListNode* node1 = new ListNode(1); head->next = node1; ListNode* node2 = new ListNode(2); node1->next = node2; ListNode* node3 = new ListNode(4); node2->next = node3; ListNode* node4 = new ListNode(5); node3->next = node4; return head; }//initList2 //从头至尾遍历单链表 void visitList(ListNode* head){ while(head!= NULL){ cout<<head->val<<endl; head= head->next; } }//visitList int main() { //函数中初始化 cout<<"第一个链表:"<<endl; ListNode* head1 = new ListNode(0); head1 = initList1(head1); visitList(head1); cout<<"第二个链表:"<<endl; ListNode* head2 = new ListNode(-1); head2 = initList2(head2); visitList(head2); cout<<"合并后的新链表:"<<endl; Solution solution;//正确 ListNode* head = solution.mergeTwoLists(head1,head2);//调用成员方法 visitList(head); return 0; } //第一个链表: //0 //2 //3 //6 //8 //第二个链表: //-1 //1 //2 //4 //5 //合并后的新链表: //-1 //0 //1 //2 //2 //3 //4 //5 //6 //8 //请按任意键继续. . .
My solve:
#include<iostream> #include<string> #include<stdio.h> #include<string.h> #include<iomanip> #include<vector> #include<list> #include<queue> #include<algorithm> #include<stack> #include<map> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* head=NULL; ListNode* next=NULL; if(l1==NULL) return l2; if(l2==NULL) return l1; if(l1->val<=l2->val) { head=l1; l1=l1->next; } else { head=l2; l2=l2->next; } next=head; while(true) { if(l1==NULL) { next->next=l2; return head; } if(l2==NULL) { next->next=l1; return head; } if(l1->val<=l2->val) { next->next=l1; next=next->next; l1=l1->next; } else { next->next=l2; next=next->next; l2=l2->next; } } return head; } }; int main() { ListNode *p,*head,*l,*head2,*l2; int n; p=(ListNode*)malloc(sizeof(ListNode*)); head=p; head2=l2; p->val=0; p->next=NULL; char ch1; while((ch1=cin.get())!='\n' ) { cin.putback(ch1); scanf("%d",&p->val); l=(ListNode*)malloc(sizeof(ListNode*)); l->val=0; l->next=NULL; p->next=l; p=p->next; } while((ch1=cin.get())!='\n' ) { cin.putback(ch1); scanf("%d",&p->val); l=(ListNode*)malloc(sizeof(ListNode*)); l->val=0; l->next=NULL; p->next=l; p=p->next; } return 0; }
In this way,you can use input to evaluate ListNode
//ll1 is the head,finally use it to return
ll1 = l1 = new ListNode(0); for (int i = 0; i < n1; i++) { l1->next = new ListNode(0); l1 = l1->next; scanf("%d", &;(l1->val)); }
22. Generate Parentheses
Medium
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Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
In this code I use a new iterator const_iterator,but in fact normal iterator can do it.Only when
we need to traverse vector we can use this const one.
In fact,when I contact DFS&BFS,I thought it was very complex,maybe it's a bit complex in thinking
But when we use recursion execute DFS,you will find it's very clear and neat.
if(in>0) means still can output "("
if(in<out) means from this time,we can output ")"
because we cannot output")" from the start,only when we get this point can we start to output“）”
#include<iostream> #include<string> #include<stdio.h> #include<string.h> #include<iomanip> #include<vector> #include<list> #include<queue> #include<algorithm> #include<stack> #include<map> using namespace std; class Solution { public: vector<string> res; vector<string> generateParenthesis(int n) { getByDFS(n,n,""); return res; } void getByDFS(int in,int out,string a) { if(in==0&&out==0) { string one=a; res.push_back(one); } else { if(in>0) { getByDFS(in-1,out,a+"("); } if(in<out) { getByDFS(in,out-1,a+")"); } } } }; int main() { Solution s; int n; vector<string> res; cin>>n; res=s.generateParenthesis(n); vector<string>::const_iterator iter=res.begin(); for(;iter!=res.end();++iter) { cout<<*iter<<endl; } return 0; }
if you use C11 this would be perfect
for(auto i : res) { // process i cout << i <<endl; // this will print all the contents of *features*
}
22. Generate Parentheses
Medium
2857
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Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
转载于:https://www.cnblogs.com/Marigolci/p/11088949.html