备战秋招day3

碎碎念：

今天有个面试，学校这里还有实训。
今日任务：算法＋mysql知识点补充

算法

24. 两两交换链表中的节点

/*** 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) return head;ListNode dump = new ListNode(-1,head);ListNode cur = dump;while(cur.next != null && cur.next.next != null){ListNode tmp = cur.next.next.next;//交换ListNode node1 = cur.next;ListNode node2 = cur.next.next;cur.next = node2;node2.next = node1;node1.next = tmp;//往后迭代cur = node1;}return dump.next;}
}

19. 删除链表的倒数第 N 个结点

/*** 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(-1,head);//快慢指针:找倒数第几个的话，那我们可以找两个指针，快指针先走n步ListNode fast = dummy;ListNode slow = dummy;int count = n;while(n-- > 0 && fast!=null){fast = fast.next;}//我们要找到待删除节点的前一个节点 fast = fast.next;while(fast!=null){slow = slow.next;fast = fast.next;}slow.next = slow.next.next;return dummy.next;}
}

面试题 02.07. 链表相交

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {//题目指的相交，是指某一段后，两条链表有公共部分，即从交点开始后半段是相同的//因此我们得找到交点，即先统计出两条表的长度public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode tmpa = headA;ListNode tmpb = headB;int lena = 0,lenb = 0;while(tmpa!=null){lena++;tmpa = tmpa.next;}while(tmpb!=null){lenb++;tmpb = tmpb.next;}//我们得对两条链表谁长做不同情况的判断//但这里有个方法可以让我们不用写两遍，就是永远让A链表是最长的那条链表if(lenb>lena){//做交换就能保证a是最长链表ListNode node = null;node = headA;headA = headB;headB = node;}//算出两者差，谁长谁先走差值的步数int diff = Math.abs((lena-lenb));//重置回来tmpa = headA;tmpb = headB;//A最长，所以A先走差值while(diff-- > 0){tmpa = tmpa.next;}//找相同节点while(tmpa!=null){//注意：题目说了是节点相同，而不仅仅是值，因此这里直接==即可if(tmpa == tmpb){return tmpa;}tmpa = tmpa.next;tmpb = tmpb.next;}return null;}
}

142. 环形链表 II

/*** Definition for singly-linked list.* class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode detectCycle(ListNode head) {ListNode fast = head;ListNode slow = head;//先判断是否有环while(fast!=null && fast.next!=null){slow = slow.next;fast = fast.next.next;//两个指针如果相遇，表示有环if(slow == fast){//找相遇点//方法是通过两个指针来去找ListNode node1 = fast;ListNode node2 = head;while(node1 != node2){node1 = node1.next;node2 = node2.next;}return node2;}}return null;}
}