标签：lin find turn highlight 今天开始 eset amp top val
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
null.好久不刷题了，一开始还真没什么思路。
这道题有多种解法。
解法一：
先求出两个linkedlist的长度，然后从长的开始，直到走到两个一样长的地方，然后对比两个从哪里开始一样。
昨天经过和斯坦福小哥的对话之后觉得真的要好好关注代码质量了，然后虽然关注了也写了一个helper function，发现还是不够好。
改良前：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) {
return null;
}
int lenA = 0;
int lenB = 0;
ListNode a = headA;
ListNode b = headB;
while (a != null) {
lenA++;
a = a.next;
}
while (b != null) {
lenB++;
b = b.next;
}
ListNode result = null;
if (lenA > lenB) {
result = helper(headA, headB, lenA - lenB);
} else {
result = helper(headB, headA, lenB - lenA);
}
return result;
}
private ListNode helper(ListNode longer, ListNode shorter, int subtraction) {
while (subtraction > 0) {
longer = longer.next;
subtraction--;
}
while (longer != null) {
if (longer != shorter) {
longer = longer.next;
shorter = shorter.next;
} else {
return longer;
}
}
return null;
}
}
改良后，好了点吧
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) {
return null;
}
int lenA = length(headA);
int lenB = length(headB);
ListNode result = null;
if (lenA > lenB) {
result = helper(headA, headB, lenA - lenB);
} else {
result = helper(headB, headA, lenB - lenA);
}
return result;
}
private ListNode helper(ListNode longer, ListNode shorter, int subtraction) {
while (subtraction > 0) {
longer = longer.next;
subtraction--;
}
while (longer != null) {
if (longer != shorter) {
longer = longer.next;
shorter = shorter.next;
} else {
return longer;
}
}
return null;
}
private int length(ListNode head) {
int len = 0;
while (head != null) {
len++;
head = head.next;
}
return len;
}
}
方法二：
利用快慢指针成环的原理，证明未完待续，以后再回顾下吧。。。linkedlist 环。
方法来自于此https://discuss.leetcode.com/topic/28067/java-solution-without-knowing-the-difference-in-len
证明在下面。
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
//boundary check
if(headA == null || headB == null) return null;
ListNode a = headA;
ListNode b = headB;
//if a & b have different len, then we will stop the loop after second iteration
while( a != b){
//for the end of first iteration, we just reset the pointer to the head of another linkedlist
a = a == null? headB : a.next;
b = b == null? headA : b.next;
}
return a;
}
今天开始也用C++刷题啦，第一道题目
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL) {
return NULL;
}
ListNode *p1 = headA;
ListNode *p2 = headB;
while (p1 != p2) {
p1 = p1 == NULL ? headB : p1->next;
p2 = p2 == NULL ? headA : p2->next;
}
return p1;
}
};
对指针还是一知半解，慢慢来吧~
Intersection of Two Linked Lists
标签：lin find turn highlight 今天开始 eset amp top val
原文地址：http://www.cnblogs.com/aprilyang/p/6939421.html
