Intersection of Two Linked Lists

• Intersection of Two Linked Lists
  • 题目描述
  • 解法

Intersection of Two Linked Lists

题目描述

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.

解法

代码如下:

public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    ListNode currA = headA, currB = headB;
    // 获取链表A和B的长度
    int lenA = 0, lenB = 0;
    while( currA != null ) { ++lenA; currA = currA.next; }
    while( currB != null ) { ++lenB; currB = currB.next; }
    // 让指针currA和currB指向平等的起始点
    currA = headA;
    currB = headB;
    if( lenA < lenB )
        while( lenB-- != lenA ) currB = currB.next;
    else
        while( lenA-- != lenB ) currA = currA.next;

    while( currA != currB ) {
        currA = currA.next;
        currB = currB.next;
    }
    return currA;
}

思考过程: 如果两个链表的长度是一样的, 那么就可以使用两个指针一对一对的比较了, 但是链表的长度可能是不一样的, 所以可以想办法让两个指针指向的起始点是平等的, 也就是两个指针的起始点到终点的长度是相等的.

所以先计算两个链表的长度, 然后让较长链表的指针先移动一段距离, 使得和另一个指针处在平等的起始点, 最后一对一对比较.

时空复杂度: 时间复杂度是O(n), 空间复杂度是O(1)