题目描述
输入两个链表，找出它们的第一个公共结点。
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
解法一：
/*
关键：找出2个链表的长度，然后让长的先走两个链表的长度差，然后再一起走
（因为2个链表用公共的尾部NULL）
*/
class Solution {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
ListNode *p1=pHead1;
ListNode *p2=pHead2;
int len1=0,len2=0,diff=0;
while(p1!=NULL){
p1=p1->next;
len1++;
}
while(p2!=NULL){
p2=p2->next;
len2++;
}
if(len1>len2){
diff=len1-len2;
p1=pHead1;
p2=pHead2;
}
else{
diff=len2-len1;
p1=pHead2;
p2=pHead1;
}
for(int i=0;i<diff;i++){
p1=p1->next;
}
while(p1!=NULL && p2!=NULL){
if(p1==p2)
break;
p1=p1->next;
p2=p2->next;
}
return p1;
}
};
解法2：相关子函数进行封装
class Solution {
public:
ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) {
int len1 = findListLenth(pHead1);
int len2 = findListLenth(pHead2);
if(len1 > len2){
pHead1 = walkStep(pHead1,len1 - len2);
}else{
pHead2 = walkStep(pHead2,len2 - len1);
}
while(pHead1 != NULL){
if(pHead1 == pHead2) return pHead1;
pHead1 = pHead1->next;
pHead2 = pHead2->next;
}
return NULL;
}
int findListLenth(ListNode *pHead1){
if(pHead1 == NULL) return 0;
int sum = 1;
while(pHead1 = pHead1->next) sum++;
return sum;
}
ListNode* walkStep(ListNode *pHead1, int step){
while(step--){
pHead1 = pHead1->next;
}
return pHead1;
}
};
