Definition for singly-linked list
比方说有链表1:n1 n2 n3 n4
有链表2:n5 n3 n4
实际上n3和n4是同一个，也就是他们在n3时交叉的
如果是分别遍历，复杂度是O(n2)，接下来介绍O(n)解法
假设长的链表为len1，短的为len2
那么把长的移动len1-len2
然后两个链表指针可以同时移动了，找到交叉的节点就返回，复杂度O(n)
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| #include <iostream>
using namespace std;
struct Node{
int val;
Node* next;
};
int getLen(Node *p){
int len = 0;
while(p != NULL){
len++;
p = p->next;
}
return len;
}
Node* listMove(Node*p, int diff){
for(int i = 0; i < diff; i++){
p = p->next;
}
return p;
}
Node* findCommon(Node *l1, Node *l2){
int len1 = getLen(l1);
int len2 = getLen(l2);
Node *longListP, *shortListP;
if(len1 > len2){
longListP = l1;
shortListP = l2;
} else{
longListP = l2;
shortListP = l1;
}
int diff = abs(len1 - len2);
longListP = listMove(longListP, diff);
while(longListP != NULL){
if(longListP == shortListP){
return longListP;
}
longListP = longListP->next;
shortListP = shortListP->next;
}
return NULL;
}
int main() {
Node n1, n2, n3, n4, n5;
Node *l1 = &n1, *l2 = &n5;
n1.val = 1;
n2.val = 2;
n3.val = 3;
n4.val = 4;
n5.val = 5;
n1.next = &n2;
n2.next = &n3;
n3.next = &n4;
n5.next = &n3;
cout << findCommon(l1, l2)->val << endl;
}
|
