給定兩個鏈表,如果這兩個鏈表相交,則返回第一個相交的頂點,如果不相交,則返回空。比如
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
則返回c1節點。
方法一:(1)求出鏈表l1和l2的長度。(2)較長的鏈表先走|l1-l2|步。(3)設置兩個指針,直到兩個指針相等時,返回結果。
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
{
if (headA == NULL || headB == NULL)
return NULL;
int len1 = 1;
int len2 = 1;
ListNode *p1 = headA;
ListNode *p2 = headB;
while (p1->next)
{
len1++;
p1 = p1->next;
}
while (p2->next)
{
len2++;
p2 = p2->next;
}
if (p1 != p2)
return NULL;
p1 = headA;
p2 = headB;
if (len1 > len2)
{
int num = len1 - len2;
while (num-- > 0)
{
p1 = p1->next;
}
}
else
{
int num = len2 - len1;
while (num-- > 0)
{
p2 = p2->next;
}
}
while (p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
return p1;
}
方法二:(1)設置兩個指針,判斷兩個指針是否爲空,且是否相等。(2)兩個指針每次都走一步,當l1指針爲空時,則把l1設置爲鏈表2的頭指針,l2指針爲空時,則把l2指針設置爲鏈表1的頭指針。
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
{
if(headA == NULL || headB == NULL)
return NULL;
ListNode *p1 = headA;
ListNode *p2 = headB;
while(p1 != NULL && p2 != NULL && p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
if(p1 == p2)
return p1;
if(p1 == NULL) p1 = headB;
if(p2 == NULL) p2 = headA;
}
return p1;
}