LeetCode 160 - Intersection of Two Linked Lists

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.

 

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    if(headA == null || headB == null) return null;
    ListNode a = headA, b = headB;
    int m = 0;
    while(a != null) {
        m++;
        a = a.next;
    }
    int n=0;
    while(b != null) {
        n++;
        b = b.next;
    }
    a = headA;
    b = headB;
    while(m>n) {
        m--;
        a = a.next;
    }
    while(n>m) {
        n--;
        b = b.next;
    }
    while(a != null) {
        if(a == b) return a;
        a = a.next;
        b = b.next;
    }
    return null;
}

 

重构了一下代码：

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    int lenA = length(headA), lenB = length(headB);
    if(lenA > lenB) return getIntersectionNode(headA, lenA, headB, lenB);
    return getIntersectionNode(headB, lenB, headA, lenA);
}
private int length(ListNode head) {
    int len = 0;
    while(head != null) {
        len++; head = head.next;
    }
    return len;
}
private ListNode getIntersectionNode(ListNode a, int m, ListNode b, int n) {
    for(int i=0; i<m-n; i++) 
        a = a.next;
    while(a != null && b != null) {
        if(a == b) return a;
        a = a.next;
        b = b.next;
    }
    return null;
}