LeetCode160. Intersection of Two Linked Lists

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本文介绍三种高效算法来确定两个单链表开始相交的节点。包括使用HashSet记录节点、通过长度差调整同步遍历及循环遍历切换链表等方法。确保O(n)时间复杂度与O(1)空间复杂度。

题目

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.

答案1：

HashSet。先把一条链放入HashSet，在从另一条链起点开始遍历，第一个重复节点就是交汇点。

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    if (headA == null || headB == null) return null;
    HashSet<ListNode> set = new HashSet<ListNode>();
    ListNode p = headA;
    do {
        set.add(p);
        p = p.next;
    } while (p != null);

    p = headB;
    while (p != null && !set.contains(p)) {
        p = p.next;
    }
    return p;
}

答案2：

先求得每条链的长度。将长的链next至和短链相同长度，然后“齐头并进”。

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    if (headA == null || headB == null) return null;
    ListNode a = headA, b = headB;
    int numA = 1, numB = 1;
    while (a.next != null) {
        numA++;
        a = a.next;
    }
    while (b.next != null) {
        numB++;
        b = b.next;
    }
    if (a != b) return null;

    a = headA;
    b = headB;
    while (numA > numB) {
        a = a.next;
        numA--;
    }
    while (numB > numA) {
        b = b.next;
        numB--;
    }

    while (a != b) {
        a = a.next;
        b = b.next;
    }

    return a;
}

答案三：

一个指针从A开头遍历至结尾，转B；另一个指针从B开头遍历至结尾，转A。len(A) + len(B) == len(B) + len(A)，从而保证到达交汇点前“齐头并进”。

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    if (headA == null || headB == null) return null;
    ListNode a = headA, b = headB;
    //关键是要保持“齐头并进”
    while (a != b) {
        a = (a == null ? headB : a.next);
        b = (b == null ? headA : b.next);
    }
    return a;
}