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LeetCode 136. Single Number.html
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<h2 id="136-Single-Number（只出现一次的数字）"><a href="#136-Single-Number（只出现一次的数字）" class="headerlink" title="136. Single Number（只出现一次的数字）"></a>136. Single Number（只出现一次的数字）</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/single-number" target="_blank" rel="noopener">https://leetcode-cn.com/problems/single-number</a></p>
<h3 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h3><p>给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。</p>
<p>说明：</p>
<p>你的算法应该具有线性时间复杂度。 你可以不使用额外空间来实现吗？</p>
<p>示例 1:</p>
<p>输入: [2,2,1]<br>输出: 1<br>示例 2:</p>
<p>输入: [4,1,2,1,2]<br>输出: 4</p>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>这题思路有点强的，我本来打算遍历加哈希来检测，之后发现了，这题可以用位运算（我从来没有想过这个问题）。<br>异或计算，如果二者相同则结果为0，不同的话结果为1，再加上异或自带的交换律，最后结果就是0异或单一数，就是要找的数字。</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></pre></td><td class="code"><pre><span class="line">public static int singleNumber(int[] nums) {</span><br><span class="line"> int n = 0;</span><br><span class="line"> for (int i = 0; i < nums.length; i++) {</span><br><span class="line"> n = n ^ nums[i];</span><br><span class="line"> }</span><br><span class="line"> return n;</span><br><span class="line"> }</span><br></pre></td></tr></table></figure>
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