代码随想录算法训练营第四天 | 链表题目 两两交换链表中的节点 删除链表的倒数第N个节点 链表相交 环形链表II

两两交换链表中的节点

涉及修改链表结构的题，定义一个虚拟头节点会更方便。

/*
 * @lc app=leetcode.cn id=24 lang=java
 *
 * [24] 两两交换链表中的节点
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode();
        ListNode cur = new ListNode();
        dummy.next = head;
        cur = dummy;
        ListNode temp = new ListNode();
        ListNode before = new ListNode();
        ListNode after = new ListNode();
        
        // temp = head.next;
        while (cur.next != null && cur.next.next != null) {
            before = cur.next.next;
            after = cur.next;
            temp = before.next;
            cur.next = before;
            before.next = after;
            after.next = temp;
            cur = after;
        }
        return dummy.next;
    }
}
// @lc code=end

删除链表的倒数第N个节点

类似数组快慢指针做法

/*
 * @lc app=leetcode.cn id=19 lang=java
 *
 * [19] 删除链表的倒数第 N 个结点
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode();
        dummy.next = head;
        ListNode first = dummy;
        ListNode second = head;
        for (int i = 0; i < n - 1; i++) {
            second = second.next;
        }
        while (second.next != null) {
            second = second.next;
            first = first.next;
        }

        first.next = first.next.next;

        return dummy.next;
    }
}
// @lc code=end

链表相交

先求出链表长度，求其差值，让长链表到剩余长度与短链表长度相等的节点，然后开始寻找相交节点

/*
 * @lc app=leetcode.cn id=160 lang=java
 *
 * [160] 相交链表
 */

// @lc code=start
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode A = headA;
        ListNode B = headB;
        int length_A = 0;
        int length_B = 0;
        while (A.next != null) {
            length_A++;
            A = A.next;
        }
        while (B.next != null) {
            length_B++;
            B = B.next;
        }
        int dif = 0;
        // System.out.printf("%d, %d", length_A, length_B);
        ListNode long_ = new ListNode();
        ListNode short_ = new ListNode();
        if (length_A > length_B) {
            long_ = headA;
            short_ = headB;
            dif = length_A - length_B;
        } else {
            long_ = headB;
            short_ = headA;
            dif = length_B - length_A;
        }
        for (int i = 0; i < dif; i++) {
            long_ = long_.next;
        }
        while (long_ != null) {
            if (long_ == short_) {
                return long_;
            }
            long_ = long_.next;
            short_ = short_.next;
        }
        return null;

    }
}
// @lc code=end

环形链表II

本质是数学题，初见没有思路，也是采用快慢指针的做法，如果指针有相遇则存在环
还需要理解，如何找到入口的公式