24. Swap Nodes in Pairs

这道题意思是，给出一个链表，把每一对相邻节点交换，并且不允许改变节点内部的val值，空间复杂度是常量级。

我定义了三个指针abc，b和c指向要交换的两个节点，a指向要交换的那一对节点的前面一个节点。每次交换b和c，然后把a更新到b（注意交换完了之后，后面的节点变成b而不是c了）

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* swapPairs(ListNode* head) {        if(head == NULL)    return NULL;        if(head->next == NULL)  return head;        ListNode* newhead = new ListNode(-1);        newhead->next = head;        ListNode* a = newhead;        ListNode* b = newhead->next;        ListNode* c = newhead->next->next;        while(a != NULL && b != NULL && c != NULL){            cout << a->val << " "<< b->val << " "<< c->val << endl;            a->next = c;            b->next = c->next;            c->next = b;            a = b;            b = a->next;            c = (b == NULL)?NULL:b->next;        }        return newhead->next;    }};

看到有更简单的代码，方法一样的：

class Solution {public:    ListNode* swapPairs(ListNode* head) {        ListNode *dummy = new ListNode(-1), *pre = dummy;        dummy->next = head;        while (pre->next && pre->next->next) {            ListNode *t = pre->next->next;            pre->next->next = t->next;            t->next = pre->next;            pre->next = t;            pre = t->next;        }        return dummy->next;    }};

还有用递归做的，更简单：

class Solution {public:    ListNode* swapPairs(ListNode* head) {        if (!head || !head->next) return head;        ListNode *t = head->next;        head->next = swapPairs(head->next->next);        t->next = head;        return t;    }};