leetcode算法之链表

1.两数相加

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* newhead = new ListNode(0);
        ListNode* prev = newhead;
        int t = 0;
        ListNode* cur1 = l1,*cur2 = l2;
        while(cur1 || cur2 || t)
        {
            if(cur1)
            {
                t += cur1-&gt;val;
                cur1 = cur1-&gt;next;
            }
            if(cur2)
            {
                t += cur2->val;
                cur2 = cur2->next;
            }
            prev->next = new ListNode(t%10);
            prev = prev->next;
            t /= 10;
        }
        prev = newhead->next;
        delete newhead;
        return prev;
    }
};

2.两两交换链表中的节点

两两交换链表中的节点
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    //模拟 循环 迭代
    ListNode* swapPairs(ListNode* head) {
        if(head == nullptr || head->next == nullptr) return head;
        ListNode* ret = new ListNode(0);
        ListNode* prev = ret;
        prev->next = head;
        ListNode* cur = prev->next,*next = cur->next,*nnext = next->next;
        while(cur &amp;&amp; next)
        {
            //交换节点
            prev->next = next;
            next->next = cur;
            cur->next = nnext;
            //更新节点
            prev = cur;
            cur = nnext;
            if(cur) next = cur->next;
            if(next) nnext = next->next;
        }
        cur = ret->next;
        delete ret;
        return cur;
    }
};

3.重排链表

重排链表
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        if(head == nullptr || head->next == nullptr || head->next->next == nullptr) return;
        ListNode* ret = new ListNode(0);
        ListNode* prev = ret;
        //1.找中间节点
        ListNode* slow = head,*fast = head;
        while(fast &amp;&amp; fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        //2.将slow后面的链表部分逆序
        ListNode* newhead = new ListNode(0);
        ListNode* cur = slow->next;
        slow->next = nullptr;//将前一段链表和后一段链表断开
        while(cur)
        {
            ListNode* next = cur->next;
            cur->next = newhead->next;
            newhead->next = cur;
            cur = next;
        }
        //3.合并两个链表
        ListNode* cur1 = head,*cur2 = newhead->next;
        while(cur1 || cur2)
        {
            if(cur1)
            {
                prev->next = cur1;
                cur1 = cur1->next;
                prev = prev->next;
            }
            if(cur2)
            {
                prev->next = cur2;
                cur2 = cur2->next;
                prev = prev->next;
            }
        }
        delete ret;
        delete newhead;
    }
};

4.合并K个升序 链表

合并K个升序链表
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {//法一：
    struct cmp
    {
        bool operator()(const ListNode* l1,const ListNode* l2)
        {
            return l1->val>l2->val;
        }
    };
public:
    ListNode* mergeKLists(vector<ListNode*>&amp; lists) {
        if(lists.size() == 0) return nullptr;
        if(lists.size() == 1) return lists[0];
        //使用优先级队列，即最小堆来解决
        priority_queue<ListNode*,vector<ListNode*>,cmp> heap;
        ListNode* ret = new ListNode(0);
        ListNode* prev = ret;
        for(auto l:lists)
        {
            if(l) heap.push(l);
        }
        while(!heap.empty())
        {
            ListNode* t = heap.top();
            heap.pop();
            prev->next = t;
            prev = t;
            if(t->next) heap.push(t->next);
        }
        prev = ret->next;
        delete ret;
        return prev;
    }
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {//法二：使用归并-递归来解决
public:
    ListNode* mergeKLists(vector<ListNode*>&amp; lists) {
        return merge(lists,0,lists.size()-1);
    }
    ListNode* merge(vector<ListNode*>&amp; lists,int left,int right)
    {
        if(left > right) return nullptr;
        if(left == right) return lists[left];
        //1.选择中间元素划分区间
        int mid = (left+right)>>1;
        //[left,mid][mid+1,right]
        //2.处理左右区间
        ListNode* l1 = merge(lists,left,mid);
        ListNode* l2 = merge(lists,mid+1,right);
        //3.合并两个有序链表
        ListNode* ret = new ListNode(0);
        ListNode* prev = ret;
        ListNode* cur1 = l1,*cur2 = l2;
        while(cur1 &amp;&amp; cur2)
        {
            if(cur1->val <= cur2->val)
            {
                prev->next = cur1;
                prev = prev->next;
                cur1 = cur1->next;
            }
            else
            {
                prev->next = cur2;
                prev = prev->next;
                cur2 = cur2->next;
            }
        }
        if(cur1) prev->next = cur1;
        if(cur2) prev->next = cur2;

        prev = ret->next;
        delete ret;
        return prev;
    }
};

5.K个一组翻转链表

K个一组翻转链表
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        //模拟
        if(head==nullptr || head->next==nullptr) return head;
        //1.计算需要翻转的组数n
        int n = 0;
        ListNode* cur = head;
        while(cur)
        {
            n++;
            cur = cur->next;
        }
        n /= k;
        cur = head;
        //2.重复n次，长度为n的链表逆序
        ListNode* ret = new ListNode(0);
        ListNode* prev = ret;
        for(int i = 0;i<n;i++)
        {
            ListNode* tmp = cur;
            for(int j = 0;j<k;j++)
            {
                ListNode* next = cur->next;
                cur->next = prev->next;
                prev->next = cur;
                cur = next;
            }
            prev = tmp;
        }
        prev->next = cur;
        prev = ret->next;
        delete ret;
        return prev;
    }
};