力扣面试150题--环形链表和两数相加

Day 32

题目描述

思路

采取快慢指针

/*** Definition for singly-linked list.* class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public boolean hasCycle(ListNode head) {ListNode fast = head;ListNode slow = head;// 空链表、单节点链表一定不会有环while (fast != null && fast.next != null) {fast = fast.next.next; // 快指针，一次移动两步slow = slow.next;      // 慢指针，一次移动一步if (fast == slow) {   // 快慢指针相遇，表明有环return true;}}return false; // 正常走到链表末尾，表明没有环}
}

题目描述

思路

由于l1和l2是倒序排列的，也就是从低位到高位，那么直接按照顺序相加，记录进位，如果l1和l2其中一个加到尾部了，另外一个加0和进位即可，最后返回结果

/*** 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 addTwoNumbers(ListNode l1, ListNode l2) {ListNode res=new ListNode();ListNode head=res;int num=0;//模拟进位while(l1!=null&&l2!=null){int x=l1.val+l2.val+num;if(x>=10){num=1;res.next=new ListNode(x%10);res=res.next;}else{num=0;res.next=new ListNode(x);res=res.next;}l1=l1.next;l2=l2.next;}if(l1==null){while(l2!=null){int x=num+l2.val;if(x>=10){num=1;res.next=new ListNode(x%10);res=res.next;}else{num=0;res.next=new ListNode(x);res=res.next;}l2=l2.next;}}else{while(l1!=null){int x=num+l1.val;if(x>=10){num=1;res.next=new ListNode(x%10);res=res.next;}else{num=0;res.next=new ListNode(x);res=res.next;}l1=l1.next;}}if(num==1){res.next=new ListNode(1);res=res.next;}return head.next;}
}