leetcode之链表类之链表归并类-----OJ 2/21/23/445 链表相加求和 链表归并

1、OJ21两个链表合并：两个链表合并可以简单的依次比较完成。

OJ21代码：

    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        if (!l1 && !l2) {            return nullptr;        } else if (!l1 || !l2) {            return l1?l1:l2;        }                ListNode *l3 = nullptr, *tail = l3;                while (l1 && l2) {            int curval = (l1->val < l2->val)?l1->val:l2->val;            if (l1->val < l2->val) {                l1 = l1->next;            } else {                l2 = l2->next;            }            if (!l3) {                l3 = new ListNode(curval);                tail = l3;            } else {                ListNode *newnode = new ListNode(curval);                tail->next = newnode;                tail = newnode;            }        }                while (l1) {            int curval = l1->val;            ListNode *newnode = new ListNode(curval);            tail->next = newnode;            tail = newnode;            l1 = l1->next;        }        while (l2) {            int curval = l2->val;            ListNode *newnode = new ListNode(curval);            tail->next = newnode;            tail = newnode;            l2 = l2->next;        }                return l3;    }

2、OJ23多个链表合并

核心思路是：将多个链表加入到小顶堆，然后重载括号运算符重载(用于获取链表value并将较大取出，相当于比较运算符，这里要注意重载的方式)，每次取出堆顶，头节点将作为新链表的当前需加入的节点。

注意：1、小顶堆的每个元素是链表，因为每个链表已经有序，且每次堆调整会把头节点更小的链表调整到堆顶，故每次从取出的堆顶链表，取其头节点放入新链表即可，

   2、然后需要把这个链表去掉头节点的部分重新放入堆，以再次调整，找到头节点更小的链表，调整到堆顶。如果该链表已空，则无需加入。

   3、这样直到全部链表都已空，即结束。

OJ23代码：

    struct cmp {        bool operator() (ListNode *a, ListNode *b) {            return a->val > b->val;        }    };        ListNode* mergeKLists(vector<ListNode*>& lists) {        std::priority_queue<ListNode *, vector<ListNode *>, cmp> hp;        for (auto l:lists) {            if (l) {                hp.push(l);            }        }                ListNode *res = nullptr, *tail = res;        if (hp.empty()) {            return res;        }        while (!hp.empty()) {            ListNode *top = hp.top();            hp.pop();                        if (!res) {                res = top;                tail = res;            } else {                tail->next = top;                tail = tail->next;            }                        if (top->next) {                hp.push(top->next);            }        }                return res;    }

3、OJ2 Add Two Numbers 和 OJ445 Add Two Numbers II

首先OJ2，如同两个链表的归并排序，两个链表从头开始value相加，注意进位的处理，直到只要有一个链表为空为止，然后对可能存在的不为空的链表继续累加，注意最后如果还有进位，还要追加一个value为1的节点；

OJ2代码：

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {                if (!l1 && !l2) {            return nullptr;        } else if (!l1 || !l2) {            return (l1)?l1:l2;        }                ListNode *l3 = nullptr, *tail = nullptr;        int add = 0;        while (l1 && l2) {            int v1 = l1->val, v2 = l2->val;            int res = v1 + v2 + add;            if (res >= 10) {                add = 1;                res = res - 10;            } else {                add = 0;            }            if (!l3) {                l3 = new ListNode(res);                tail = l3;            } else {                ListNode *newnode = new ListNode(res);                tail->next = newnode;                tail = newnode;            }            l1 = l1->next;            l2 = l2->next;        }                while (l1) {            int v = l1->val;            int res = v + add;            if (res >= 10) {                add = 1;                res = res - 10;            } else {                add = 0;            }            ListNode *newnode = new ListNode(res);            tail->next = newnode;            tail = newnode;            l1 = l1->next;        }        while (l2) {            int v = l2->val;            int res = v + add;            if (res >= 10) {                add = 1;                res = res - 10;            } else {                add = 0;            }            ListNode *newnode = new ListNode(res);            tail->next = newnode;            tail = newnode;            l2 = l2->next;        }                if (add) {            ListNode *newnode = new ListNode(1);            tail->next = newnode;        }                return l3;    }};

对于OJ445，区别是需要先做原链表的翻转，然后再做OJ2的流程，也就是说数的顺序是反过来的

OJ445代码：

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        if (!l1 && !l2) {            return nullptr;        } else if (!l1 || !l2) {            return l1?l1:l2;        }                ListNode *cur = l1, *prev = nullptr, *next = l1->next;        while (next) {            ListNode *nn = next->next;            next->next = cur;            cur->next = prev;            prev = cur;            cur = next;            next = nn;        }        ListNode *rl1 = cur;                cur = l2;        prev = nullptr;        next = l2->next;        while (next) {            ListNode *nn = next->next;            next->next = cur;            cur->next = prev;            prev = cur;            cur = next;            next = nn;        }        ListNode *rl2 = cur;                ListNode *rl3 = nullptr, *tail = nullptr;        int add = 0;        while (rl1 && rl2) {            int v = rl1->val + rl2->val + add;            if (v >= 10) {                add = 1;                v = v - 10;            } else {                add = 0;            }                        if (!rl3) {                rl3 = new ListNode(v);                tail = rl3;            } else {                ListNode *newnode = new ListNode(v);                tail->next = newnode;                tail = newnode;            }            rl1 = rl1->next;            rl2 = rl2->next;        }                while (rl1) {            int v = rl1->val + add;            if (v >= 10) {                add = 1;                v = v - 10;            } else {                add = 0;            }                        ListNode *newnode = new ListNode(v);            tail->next = newnode;            tail = newnode;            rl1 = rl1->next;        }        while (rl2) {            int v = rl2->val + add;            if (v >= 10) {                add = 1;                v = v - 10;            } else {                add = 0;            }                        ListNode *newnode = new ListNode(v);            tail->next = newnode;            tail = newnode;            rl2 = rl2->next;        }                if (add) {            tail->next = new ListNode(1);        }                cur = rl3;        prev = nullptr;        next = cur->next;        while (next) {            ListNode *nn = next->next;            next->next = cur;            cur->next = prev;            prev = cur;            cur = next;            next = nn;        }                return cur;    }};