leetcode:Intersection of Two Linked Lists(两个链表的交叉点)
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:
If the two linked lists have no intersection at all, return null
.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.
分析:题意即为寻找两个单链表相交开始的节点
如果我们直接通过遍历两个链表来寻找交叉元点
代码如下:
/**
* 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 *result=NULL;
while(headB){
if(headA && headA->val == headB->val) {
result = headA;
}
while (headA)
{
if (headA->next && headA->next->val == headB->val) {
result=headA->next;
}
headA = headA->next;
}
if(headB->next) headB= headB->next;
} return result;
}
};
会出现超时:Time Limit Exceeded
看来还是得认真审题加分析,找到解题的关键特征来提高效率。(一个星期没刷题就变得头脑简单了)
最简单的代码是这样子滴:
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { while (headA && headB) {
if (headA->val < headB->val)
headA = headA->next;
else if (headA->val > headB->val)
headB = headB->next;
else if (headA->val == headB->val)
return headA;
}
return nullptr;
}
};
当然也可以是这样的思路:
我们可以遍历两个链表得到各自的长度,然后将长度更长的链表向前移动,使两个链表进行对齐,之后一起遍历,直到找到第一个相同的节点。
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { auto currA = headA, currB = headB;
int countA = 0, countB = 0; while (currA) {
currA = currA->next, countA++;
}
while (currB) {
currB = currB->next, countB++;
}
int diff = std::abs(countA - countB);
if (countB > countA) { swap(headA, headB); }
while (diff--) {
headA = headA->next;
}
while (headA != headB) {
headA = headA->next, headB = headB->next;
}
return headA;
}
};
或者:
为了节省计算,在计算链表长度的时候,顺便比较一下两个链表的尾节点是否一样,若不一样,则不可能相交,直接可以返回NULL
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(headA == NULL || headB == NULL)
return NULL;
ListNode* iter1 = headA;
ListNode* iter2 = headB;
int len1 = 1;
while(iter1->next != NULL)
{
iter1 = iter1->next;
len1 ++;
}
int len2 = 1;
while(iter2->next != NULL)
{
iter2 = iter2->next;
len2 ++;
}
if(iter1 != iter2)
return NULL;
if(len1 > len2)
{
for(int i = 0; i < len1-len2; i ++)
headA = headA->next;
}
else if(len2 > len1)
{
for(int i = 0; i < len2-len1; i ++)
headB = headB->next;
}
while(headA != headB)
{
headA = headA->next;
headB = headB->next;
}
return headA;
}
};
leetcode:Intersection of Two Linked Lists(两个链表的交叉点)的更多相关文章
- [LeetCode] Intersection of Two Linked Lists 两链表是否相交
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- intersection of two linked lists.(两个链表交叉的地方)
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- lintcode 中等题:Intersection of Two Linked Lists 两个链表的交叉
题目 两个链表的交叉 请写一个程序,找到两个单链表最开始的交叉节点. 样例 下列两个链表: A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3 在节点 c1 开始交 ...
- LeetCode: Intersection of Two Linked Lists 解题报告
Intersection of Two Linked Lists Write a program to find the node at which the intersection of two s ...
- [LeetCode]160.Intersection of Two Linked Lists(2个链表的公共节点)
Intersection of Two Linked Lists Write a program to find the node at which the intersection of two s ...
- [LeetCode] Intersection of Two Linked Lists 求两个链表的交点
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
- LeetCode Intersection of Two Linked Lists
原题链接在这里:https://leetcode.com/problems/intersection-of-two-linked-lists/ 思路:1. 找到距离各自tail 相同距离的起始List ...
- LeetCode——Intersection of Two Linked Lists
Description: Write a program to find the node at which the intersection of two singly linked lists b ...
- Intersection of Two Linked Lists两链表找重合节点
Write a program to find the node at which the intersection of two singly linked lists begins. For ex ...
随机推荐
- 【bzoj1085】[SCOI2005]骑士精神
1085: [SCOI2005]骑士精神 Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 1757 Solved: 961[Submit][Statu ...
- Auto Layout - 使用总结
1.要开始使用AutoLayout,请先设置要约束的view的translatesAutoresizingMaskIntoConstraints属性为NO.在xib或者sb中勾选Use Auto La ...
- iis express 启动多个网站
iis express一次只能运行一个网站, 执行iisexpress 不加参数. 将执行配置文件中的第一个网站. iis express一次只能运行一个 应用程序池. 可以使用这个特点实现一次 ...
- request 获取请求参数
/** * 根据request获取请求的用户参数 * @return * @return */ protected <T> T getParamConvertEntity(Class cl ...
- 10个jQuery插件分享
原文:http://www.shejidaren.com/10-jquery-plugins.html blur.js blur.js是一个很有意思的插件,它能实现像WIN7 AERO效果的JS插件, ...
- CRF++中文分词使用指南
http://blog.csdn.net/marising/article/details/5769653 前段时间写了中文分词的一些记录里面提到了CRF的分词方法,近段时间又研究了一下,特把方法写下 ...
- [C++]内存字节对齐
当我们写一个class类,然后sizeof(),然后发现这个值往往比你想象的大,这是为什么呢?这里就要讲到内存对齐的问题. 先来看一下内存对齐的几条原则: 1.对于class(struct/union ...
- Javascript Arguments,calle,caller,call,apply
一.Arguments 该对象代表正在执行的函数和调用他的函数的参数. [function.]arguments[n] 参数function :选项.当前正在执行的 Function 对象的名字. n ...
- WCF分布式开发步步为赢(10):请求应答(Request-Reply)、单向操作(One-Way)、回调操作(Call Back).
WCF除了支持经典的请求应答(Request-Reply)模式外,还提供了什么操作调用模式,他们有什么不同以及我们如何在开发中使用这些操作调用模式.今天本节文章里会详细介绍.WCF分布式开发步步为赢( ...
- 收缩SQL数据库日志文件
收缩SQL数据库日志文件 介绍具体的操作方法前,先说下我操作的实际环境和当时的状况.我的服务器是windows server 2008 R2 64位英文版,数据库是SQL server 2008英文版 ...