題目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.
思路:
- 求得兩個(gè)鏈表長(zhǎng)度lenA,lenB. 這里假設(shè)lenA>lenB,
- 鏈表A前進(jìn)(lenB-lenA)個(gè)節(jié)點(diǎn),這樣和鏈表B對(duì)齊
- 鏈表A和鏈表B同時(shí)向前每次移動(dòng)一個(gè)節(jié)點(diǎn),進(jìn)行對(duì)比,如果相等返回這個(gè)節(jié)點(diǎn),如果兩者都到鏈表尾部不
相等,則返回null
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int lenA = getListLength(headA);
int lenB = getListLength(headB);
while(lenA > lenB){
headA = headA.next;
lenA--;
}
while(lenA < lenB){
headB = headB.next;
lenB--;
}
while(headA != headB){
headA = headA.next;
headB = headB.next;
}
return headA;
}
private int getListLength(ListNode head){
int len = 0;
ListNode node = head;
while(node != null){
len++;
node = node.next;
}
return len;
}
