leetcode 24. Swap Nodes in Pairs \ 21. Merge Two Sorted Lists

24. swap nodes in pairs

这个算法就是把前后节点交换为知，记住每个节点存储在内存中某个地方，指向每个节点的地址是一样，不一样的是每个节点的下一个节点的地址，改变这个即可。

法一：

/** * 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) {        ListNode *node_walker = head;        ListNode *node_next = NULL;        ListNode *node_cur = NULL;        ListNode *node_pre= NULL;        ListNode *node_next2 = NULL;        while(node_walker && node_walker->next)        {            node_next = node_walker->next;            node_next2 = node_next->next;            node_cur = node_walker;            if(node_walker == head)            {                head = node_next;                head->next = node_cur;                node_cur->next = node_next2;            }            else            {                node_pre->next = node_next;                node_walker->next = node_next2;                node_next->next = node_walker;            }            node_pre = node_walker;            node_walker = node_next2;        }        return head;    }};

法二：
上面的算法可以写得更加简洁，那就是：
这里有一个巧妙的用法就是它索引指向指针的指针，然后修改这个指针的话就会使得值的赋值过程减少一个阻碍。

/** * 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) {        ListNode **head_new = &head, *node_cur = NULL, *node_next = NULL;        while ((node_cur = *head_new) && (node_next = node_cur->next)) {            node_cur->next = node_next->next;            node_next->next = node_cur;            *head_new = node_next;            head_new = &(node_cur->next);        }        return head;    }};

21. Merge Two Sorted Lists

法一：

/** * Definition for singly-linked list. * 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 *tmp = NULL;        while(l1 != NULL && l2 != NULL)        {            if(l1->val <= l2->val)            {                if(!head)                {                    head = l1;                    tmp = head;                }                    else                {                    tmp->next = l1;                    tmp = l1;                }                    l1 = l1->next;            }            else            {                if(!head)                {                    head = l2;                    tmp = head;                }                    else                {                    tmp->next = l2;                    tmp = l2;                }                    l2 = l2->next;            }            if(!l1)            {                tmp->next = l2;                break;            }            if(!l2)            {                tmp->next = l1;                break;            }        }        head = (head == NULL)?((l1 == NULL)?l2:l1):head;        return head;    }};

法二：
精简版：

class Solution {public:    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {        ListNode dummy(INT_MIN);        ListNode *tail = &dummy;        while (l1 && l2) {            if (l1->val < l2->val) {                tail->next = l1;                l1 = l1->next;            } else {                tail->next = l2;                l2 = l2->next;            }            tail = tail->next;        }        tail->next = l1 ? l1 : l2;        return dummy.next;    }};