leetcode代码分类汇总之-链表

leetcode上链表的题目还算不少的，暂时收录下面这些，可能有些被我分到其他部分去了。

Add Two Numbers

code:

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int carry=0;
        ListNode* dummy = new ListNode(0),*ite=dummy;
        while(l1&&l2){
            int num = l1->val+l2->val+carry;
            carry = num/10;
            num %=10;
            ite->next = new ListNode(num);
            ite = ite->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        l1 = l2?l2:l1;
        while(l1){
            int num = l1->val+carry;
            carry = num/10;
            num %=10;
            ite->next = new ListNode(num);
            ite=ite->next;
            l1 = l1->next;
        }
        if(carry){
            ite->next = new ListNode(carry);
            ite = ite->next;
        }
        ite->next = NULL;
        ite = dummy->next;
        delete dummy;
        return ite;
    }
};

这个题比较简单，写起来陷阱也不算多，一个改版是，如果链表的数字是倒过来的，怎么做？也就是说链表头是最高位的数字而不是最低位。
可以用递归做。

Partition List

这个是很久之前写的代码，O(n)时间+O(n)空间。应该在代码风格和效率上还有可以优化的地方，暂时放在这里，以后有好的代码再补一个。

code：

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode* p1 = head;
        if(head==NULL) return head;
        vector vec;
        while(p1!=NULL){
            vec.push_back(p1->val);
            p1 = p1->next;
        }
        p1 = head;
        bool sign = true;
        int pos = 0;
        while(p1!=NULL){
            while(sign&&pos=x)
                pos++;
            while(!sign&&posval = vec[pos++];
                p1 = p1->next;
            }else{
                if(!sign)
                    p1 = p1->next;
                sign = false;
                pos = 0;
            }
        }
        return head;
    }
};

关于这个题，还有一个很相似的版本，是要求O(n)时间+O(1)空间做到这个事情，不过给的不是链表，而是数组。我估摸着要是能做到，岂不是可以写出一个稳定的快排？？不知道怎么搞。

Remove Duplicates from Sorted List

code：

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *ite,*pre=head;
        if(!head) return head;
        ite = head->next;
        while(ite){
            if(ite->val==pre->val){
                pre->next = ite->next;
                delete ite;
            }else
                pre = ite;
            ite = pre->next;
        }
        return head;
    }
};

一开始没注意是sorted，写了个没有sorted的代码，顺便也附上：

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode* dummy = new ListNode(1),*ite,*pre=dummy;;
        dummy->next = head;
        ite = head;
        set s;
        while(ite){
            if(s.find(ite->val)!=s.end()){
                pre->next = ite->next;
                delete ite;
                ite=pre->next;
            }else{
                s.insert(ite->val);
                pre = ite;
                ite = ite->next;
            }
        }
        head = dummy->next;
        delete dummy;
        return head;
    }
};

Remove Duplicates from Sorted List II

code：

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode* dummy = new ListNode(1);
        dummy->next = head;
        bool sign = false;
        ListNode* pre = dummy,*ite=head;
        while(ite){
            if(sign||(ite->next&&ite->next->val==ite->val)){
                if(!ite->next||ite->next->val!=ite->val)
                  sign =false;
                else sign = true;
                pre->next = ite->next;
                delete ite;
            }else{
                pre=ite;
            }
            ite = pre->next;
        }
        head = dummy->next;
        delete dummy;
        return head;
    }
};

Remove Nth Node From End of List

code：

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode** ite = &head, *runner = head,*tmp;
        while(n--&&runner)
            runner = runner->next;
        if(n+1) return head;
        while(runner)
        {
            runner = runner->next;
            ite = &((*ite)->next);
        }
        tmp = *ite;
        (*ite) = (*ite)->next;
        delete tmp;
        return head;
    }
};

这个代码的思路就是让一个指针先走n步，那第一个指针到了尾之后，第二个指针就是倒数第n个，然后删掉。

另一个思路是递归，递归结束时记录当前是倒数第几个节点：

class Solution {
public:
    int DelRecursive(ListNode* head,int n){
        if(!head) return 0;
        int k = DelRecursive(head->next,n);
        if(n==k){
            ListNode* tmp = head->next;
            head->next = head->next->next;
            delete tmp;
        }
        return k+1;
    }
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        DelRecursive(dummy,n);
        head = dummy->next;
        delete dummy;
        return head;
    }
};

Reverse Linked List II

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        ListNode* dummy = new ListNode(1),*ite = dummy,*tail;
        dummy->next = head;
        for(int i=1;inext;
        head = ite;
        tail = ite->next;
        //先断开，再连接，顺序很重要,第m个不用管
        for(int i=0;inext;
            tail->next = ite->next;
            ite->next = head->next;
            head->next = ite;
        }
        head = dummy->next;
        delete dummy;
        return head;
    }
};

记住先断开再连接。

Swap Nodes in Pairs

为了体验一下指针的指针跟dummy head的区别，这个题分别用两种方法做了一遍：

2D Pointer code：

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        ListNode** ite = &head;
        while(*ite&&(*ite)->next)
        {
            ListNode* tmp = (*ite)->next;
            (*ite)->next = tmp->next;
            tmp->next = *ite;
            *ite = tmp;
            ite = &((*ite)->next->next);
        }
        return head;
    }
};

dummy head code：

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        ListNode* dummy = new ListNode(0),*ite=dummy;
        dummy->next = head;
        while(ite->next&&ite->next->next){
            ListNode* tmp = ite->next->next;//保存要拿走的节点
            ite->next->next = tmp->next;  //删除该节点
            tmp->next = ite->next;  //把该节点接到ite后面
            ite->next = tmp;
            ite = tmp->next;
        }
        ite = dummy->next;
        delete dummy;
        return ite;
    }
};