LeetCode 160:Intersection of Two Linked Lists
题意描述
查找两个单链表的交点开始的节点。
测试用例
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 Output: Reference of the node with value = 8Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 Output: Reference of the node with value = 2Example 3:
Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 Output: null
解题思路
一、思路一
- 从头开始遍历两个链表，a指针指向链表A，b指针指向链表B。
- 当遍历完本链表时，从另一条链表头部开始遍历
- 当两个指针指向同一个节点时，返回当前节点
Case 1 (Have Intersection & Same Len):
a
A: a1 → a2 → a3
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
b
a
A: a1 → a2 → a3
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
b
a
A: a1 → a2 → a3
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
b
A: a1 → a2 → a3
↘ a
c1 → c2 → c3 → null
↗ b
B: b1 → b2 → b3
Since a == b is true, end loop while(a != b), return the intersection node a = c1.
Case 2 (Have Intersection & Different Len):
a
A: a1 → a2
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
b
a
A: a1 → a2
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
b
A: a1 → a2
↘ a
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
b
A: a1 → a2
↘ a
c1 → c2 → c3 → null
↗ b
B: b1 → b2 → b3
A: a1 → a2
↘ a
c1 → c2 → c3 → null
↗ b
B: b1 → b2 → b3
A: a1 → a2
↘ a = null, then a = b1
c1 → c2 → c3 → null
↗ b
B: b1 → b2 → b3
A: a1 → a2
↘
c1 → c2 → c3 → null
↗ b = null, then b = a1
B: b1 → b2 → b3
a
b
A: a1 → a2
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
a
b
A: a1 → a2
↘
c1 → c2 → c3 → null
↗
B: b1 → b2 → b3
a
A: a1 → a2
↘ b
c1 → c2 → c3 → null
↗ a
B: b1 → b2 → b3
Since a == b is true, end loop while(a != b), return the intersection node a = c1.
Case 3 (Have No Intersection & Same Len):
a
A: a1 → a2 → a3 → null
B: b1 → b2 → b3 → null
b
a
A: a1 → a2 → a3 → null
B: b1 → b2 → b3 → null
b
a
A: a1 → a2 → a3 → null
B: b1 → b2 → b3 → null
b
a = null
A: a1 → a2 → a3 → null
B: b1 → b2 → b3 → null
b = null
Since a == b is true (both refer to null), end loop while(a != b), return a = null.
Case 4 (Have No Intersection & Different Len):
a
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
b
a
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
b
a
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
b
a
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
b = null, then b = a1
b a = null, then a = b1
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
b
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
a
b
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
a
b
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
a
b = null
A: a1 → a2 → a3 → a4 → null
B: b1 → b2 → b3 → null
a = null
Since a == b is true (both refer to null), end loop while(a != b), return a = null.
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if(headA == null || headB == null) return null;
ListNode a = headA;
ListNode b = headB;
while(a != b){
a = a==null?headB:a.next;
b = b==null?headA:b.next;
}
return a;
}
}
二、思路二
根据两条链表的长度，从后续长度相同的位置同时开始向后遍历，如果两条链表存在交点，返回交点。
public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode p1 = headA, p2 = headB;
int len1 = 0, len2 = 0;
while (p1 != null) {
p1 = p1.next;
len1++;
}
while (p2 != null) {
p2 = p2.next;
len2++;
}
p1 = headA;
p2 = headB;
if (len1 > len2) {
for (int i = 0; i < len1 - len2; i++) {
p1 = p1.next;
}
} else {
for (int i = 0; i < len2 - len1; i++) {
p2 = p2.next;
}
}
while (p1 != p2) {
p1 = p1.next;
p2 = p2.next;
}
return p1;
}
