刷题160. Intersection of Two Linked Lists

一、题目说明

题目160. Intersection of Two Linked Lists，计算两个链表相连的位置。难度是Easy！

二、我的解答

这个题目，简单思考一下还是容易的。一次遍历，找到ListA、ListB的最后一个元素及其长度，如果endA==endB则相交。先移动长链表的指针abs(numA-numB)，然后找到相等的位置即可。

代码如下：

class Solution{
	public:
		ListNode* getIntersectionNode(ListNode* headA, ListNode* headB){
			if(headA==NULL || headB==NULL) return NULL;
			ListNode* endA = headA,*endB=headB;
			int numA = 1,numB = 1;
			
			//find the last element of headA
			while(endA->next !=NULL){
				endA = endA->next;
				numA++;
			}
			
			//find the last element of headB
			while(endB->next !=NULL){
				endB = endB->next;
				numB++;
			}
			if(endA != endB){
				return NULL;
			}else{
				endA = headA;
				endB = headB;
				if(numA>numB){
					int t = numA-numB;
					while(t>0){
						endA = endA->next;
						t--;
					}
				}else if(numA<numB){
					int t = numB-numA;
					while(t>0){
						endB = endB->next;
						t--;
					}
				}
				
				while(endA != endB){
					endA = endA->next;
					endB = endB->next;
				}
				return endA;
			}
		}
};

性能如下：

Runtime: 44 ms, faster than 96.04% of C++ online submissions for Intersection of Two Linked Lists.
Memory Usage: 16.9 MB, less than 59.26% of C++ online submissions for Intersection of Two Linked Lists.

三、优化措施

可以继续优化，但意义不大。