Copy List with Random Pointer & Add Two Numbers & Valid Palindrome

(1) Copy List with Random Pointer 

根据[1]：不用保存原始链表的映射关系，构建新节点时，指针做如下变化，即把新节点插入到相应的旧节点后面：



分两步：

 

1、构建新节点random指针：new1->random = old1->random->next, new2-random = NULL, new3-random = NULL, new4->random = old4->random->next

 

2、恢复原始链表以及构建新链表：例如old1->next = old1->next->next,  new1->next = new1->next->next

class Solution {
public:
RandomListNode *copyRandomList(RandomListNode *head) {
RandomListNode *cur,*curnew,*ret;

cur=head;

if(!cur)
return cur;

while(cur)
{
curnew=new RandomListNode(cur->label);
curnew->next=cur->next;
cur->next=curnew;
cur=curnew->next;
}

cur=head;
curnew=head->next;

while(1)
{
if(cur->random)
curnew->random=cur->random->next;
if(!curnew->next)
break;
cur=cur->next->next;
curnew=cur->next;
}

cur=head;
curnew=head->next;
ret=curnew;

while(curnew->next)
{
cur->next=cur->next->next;
curnew->next=curnew->next->next;
cur=cur->next;
curnew=curnew->next;
}

cur->next=NULL;

return ret;
}
};

(2) Add Two Numbers
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
int add1,add2,extra=0;
ListNode head(0),*cur,*pre;

pre=&head;

while(l1 || l2)
{
if(l1)
{
add1=l1->val;
l1=l1->next;
}
else
add1=0;
if(l2)
{
add2=l2->val;
l2=l2->next;
}
else
add2=0;

int sum=add1+add2+extra;

cur=new ListNode(sum%10);
pre->next=cur;
pre=cur;
extra=sum/10;
}

if(extra)
{
cur=new ListNode(1);
pre->next=cur;
}

return head.next;
}
};

(3) Valid Palindrome
class Solution {
bool isStr(char &c) {
if(c<='9' && c>='0')
return true;
if(c<='z' && c>='a')
return true;
if(c<='Z' && c>='A')
{
c+=32;
return true;
}
return false;
}

public:
bool isPalindrome(string s) {
if(s.size()==0)
return true;

int i=0,j=s.size()-1;

while(i<j)
{
while(!isStr(s[i]) && i<j)
i++;
while(!isStr(s[j]) && i<j)
j--;

if(s[i]!=s[j])
return false;
i++;
j--;
}
return true;
}
};

参考：

[1] http://www.2cto.com/kf/201310/253477.html