[Leetcode]160.Intersection of Two Linked Lists

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本文介绍了一种在两个单链表中找到它们开始相交节点的方法。该算法能在O(n)时间内运行，并仅使用O(1)内存。适用于没有环的链表结构。

160 Intersection of Two Linked Lists
Easy

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.

For example, the following two linked lists:
40ms:

public int reverse(int x) {
        String num = x+"";
        int start = 0,end = num.length()-1;
        int t = 1;
        if(num.charAt(0)=='+'||num.charAt(0)=='-'){
            start++;
            if(num.charAt(0)=='-') t = -1;
        }
        Long res =(long) 0;
        while(end>=start){
            res = res*10 + (num.charAt(end)-'0');
            end--;
        }
        if(res>Integer.MAX_VALUE) return 0;
        else return (int) ((int) t*res);
}

 public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
         //boundary check
    if(headA == null || headB == null) return null;

    ListNode a = headA;
    ListNode b = headB;

    //if a & b have different len, then we will stop the loop after second iteration
    while( a != b){
        //for the end of first iteration, we just reset the pointer to the head of another linkedlist
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }

    return a; 
    }