两个链表的第一个公共结点_链表中环的入口结点

两个链表的第一个公共结点

题目链接

//先计算长度差,然后让一个指针先走差值单位!
public class Solution {
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
            ListNode cur1 = pHead1;
            ListNode cur2 = pHead2;
        int size1 = 0;
        int size2 = 0;
        while(cur1!=null){
            cur1 = cur1.next;
            size1++;
        }
        while(cur2!=null){
            cur2 = cur2.next;
            size2++;
        }
        if(size1>size2){
            int len = size1 - size2;
            while(len-->0){
                pHead1 = pHead1.next;
            }
        }else{
             int len = size2 - size1;
            while(len-->0){
                pHead2 = pHead2.next;
            }
        }
         
        while(pHead1!=null){
            if(pHead1.val==pHead2.val){
                return pHead1;
            }
            pHead1 = pHead1.next;
            pHead2 = pHead2.next;
        }
        return null;
    }
}

• 通过两个指针速度相同,走过的路程相同必会相遇!
• cur1 走完 L1,cur1指向 L2,cur2走完L2,指向L1!

public class Solution {
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
            //定义2个指针! 
            // cur1 走完 L1,又从 L2开始!
            // cur2 走完 L2,又从 L1开始!
            // 这里两个指针速度相同,走过的长度相等,如果有相同节点肯定相遇!
        ListNode cur1 = pHead1;
        ListNode cur2 = pHead2;
        while(cur1!=cur2){//不存在公共节点,两个指针会来到null相等退出循环!
            cur1 = (cur1==null) ? pHead2 : cur1.next;
            cur2 = (cur2 == null) ? pHead1 : cur2.next;
        }
        return cur1;
    }
}

链表中环的入口结点

链表中环的入口结点

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
        //快慢指针,利用链表头到入口距离 = 相遇点到入口距离!
        //所以当两个节点相遇后再走L距离就是入口位置!
        //相遇后让其中一个指针从链头开始走L,一个从相遇点开始!
        ListNode slow = pHead,fast = pHead;
        while(fast!=null&&fast.next!=null){//注意判断条件!!!!
            fast = fast.next.next;
            slow = slow.next;
            if(fast==slow){
                //相遇!
                //让slow从头结点开始!
                slow = pHead;
                while(fast!=slow){
                    slow = slow.next;
                    fast = fast.next;
                }
                return fast;
            }
        }
        return null;
    }
}