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LeetCode 160. Intersection of Two Linked Lists.html
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<h1 class="post-title" itemprop="name headline">LeetCode 160. Intersection of Two Linked Lists</h1>
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<span class="page-pv"><i class="fa fa-file-o"></i> 阅读量
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<h2 id="160-Intersection-of-Two-Linked-Lists（相交链表）"><a href="#160-Intersection-of-Two-Linked-Lists（相交链表）" class="headerlink" title="160. Intersection of Two Linked Lists（相交链表）"></a>160. Intersection of Two Linked Lists（相交链表）</h2><h3 id="链接"><a href="#链接" class="headerlink" title="链接"></a>链接</h3><p><a href="https://leetcode-cn.com/problems/intersection-of-two-linked-lists" target="_blank" rel="noopener">https://leetcode-cn.com/problems/intersection-of-two-linked-lists</a></p>
<h3 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h3><p>编写一个程序，找到两个单链表相交的起始节点。</p>
<p>示例 1：</p>
<p> 输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3<br> 输出：Reference of the node with value = 8<br> 输入解释：相交节点的值为 8 （注意，如果两个列表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [4,1,8,4,5]，链表 B 为 [5,0,1,8,4,5]。在 A 中，相交节点前有 2 个节点；在 B 中，相交节点前有 3 个节点。</p>
<p>示例 2：</p>
<p> 输入：intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1<br> 输出：Reference of the node with value = 2<br> 输入解释：相交节点的值为 2 （注意，如果两个列表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [0,9,1,2,4]，链表 B 为 [3,2,4]。在 A 中，相交节点前有 3 个节点；在 B 中，相交节点前有 1 个节点。</p>
<p>示例 3：</p>
<p> 输入：intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2<br> 输出：null<br> 输入解释：从各自的表头开始算起，链表 A 为 [2,6,4]，链表 B 为 [1,5]。由于这两个链表不相交，所以 intersectVal 必须为 0，而 skipA 和 skipB 可以是任意值。<br> 解释：这两个链表不相交，因此返回 null。</p>
<p>注意：</p>
<p> 如果两个链表没有交点，返回 null.<br> 在返回结果后，两个链表仍须保持原有的结构。<br> 可假定整个链表结构中没有循环。<br> 程序尽量满足 O(n) 时间复杂度，且仅用 O(1) 内存。</p>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>假设两个链表长度为a+x和b+x，x就是相同部分，那么首先两个结点都从头结点开始，当自己的链表遍历完后跳到另外一个链表头，当每个节点都走a+b+x步之后，两个链表应该相遇在交点处。若不相交，最后都是null。</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line">public class ListNode {</span><br><span class="line"></span><br><span class="line"> int val;</span><br><span class="line"> ListNode next;</span><br><span class="line"></span><br><span class="line"> ListNode(int x) {</span><br><span class="line"> val = x;</span><br><span class="line"> next = null;</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line">public ListNode getIntersectionNode(ListNode headA, ListNode headB) {</span><br><span class="line"></span><br><span class="line"> if (headA == null || headB == null) {</span><br><span class="line"> return null;</span><br><span class="line"> }</span><br><span class="line"> ListNode pA = headA, pB = headB;</span><br><span class="line"> while (pA != pB) {</span><br><span class="line"> if (pA == null) {</span><br><span class="line"> pA = headB;</span><br><span class="line"> } else {</span><br><span class="line"> pA = pA.next;</span><br><span class="line"> }</span><br><span class="line"> if (pB == null) {</span><br><span class="line"> pB = headA;</span><br><span class="line"> } else {</span><br><span class="line"> pB = pB.next;</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> }</span><br><span class="line"> return pA;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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