数据结构算法合集——链表篇

day01 剑指 Offer 24. 反转链表

双指针法

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function(head) {
//使用一个cur节点保存head节点
    let cur = head
    let pre = null
    while(cur) {
    //使用一个next节点保存cur节点的下一个节点
    //因为接下来要改变cur节点的指针指向了
        let next = cur.next
        //这步骤就相当于翻转操作
        cur.next = pre
        //移动pre节点
        pre = cur
        //移动cur节点
        cur = next
    }
    return pre
};
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递归法

和双指针法思路是一样的，只是写法有些不同

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverse(ListNode* cur, ListNode* pre) {
        if(!cur) {
            return pre;
        }
        ListNode* temp = cur->next;
        cur->next = pre;
        return reverse(temp, cur);
    }
    ListNode* reverseList(ListNode* head) {
        return reverse(head, NULL);
    }
};
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24. 两两交换链表中的节点

设置一个dummy节点直接模拟

/**
 * 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* swapPairs(ListNode* head) {
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* cur = dummyHead;
        while(cur->next != nullptr && cur->next->next != nullptr) {
            ListNode* t1 = cur->next;
            ListNode* t2 = cur->next->next->next;
            cur->next = cur->next->next;
            cur->next->next = t1;
            cur->next->next->next = t2;
            cur = cur->next->next;
        }
        return dummyHead->next;
    }
};
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19. 删除链表的倒数第 N 个结点

快慢指针法

/**
 * 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* removeNthFromEnd(ListNode* head, int n) {
       ListNode* dummyHead = new ListNode(0);
       dummyHead->next = head;
       ListNode* slow = dummyHead;
       ListNode* fast = dummyHead;
       while(n-- && fast != NULL) {
           fast = fast->next;
       }
       fast = fast->next;
       while(fast) {
           fast = fast->next;
           slow = slow->next;
       }
       slow->next = slow->next->next;
       return dummyHead->next;
    }
};
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