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LeetCode 448. Find All Numbers Disappeared in an Array.html
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<h2 id="448-Find-All-Numbers-Disappeared-in-an-Array（找到所有数组中消失的数字）"><a href="#448-Find-All-Numbers-Disappeared-in-an-Array（找到所有数组中消失的数字）" class="headerlink" title="448. Find All Numbers Disappeared in an Array（找到所有数组中消失的数字）"></a>448. Find All Numbers Disappeared in an Array（找到所有数组中消失的数字）</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/find-all-numbers-disappeared-in-an-array" target="_blank" rel="noopener">https://leetcode-cn.com/problems/find-all-numbers-disappeared-in-an-array</a></p>
<h3 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h3><p>给定一个范围在 1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次，另一些只出现一次。</p>
<p>找到所有在 [1, n] 范围之间没有出现在数组中的数字。</p>
<p>您能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内。</p>
<p>示例:</p>
<p> 输入:<br> [4,3,2,7,8,2,3,1]</p>
<p> 输出:<br> [5,6]</p>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>这题的题意很简单，就是要找到缺失的数字，但是思路很巧妙，对于每个数字，都设置一个位置，就像例子一样，第一个数字为4，就将第四个数字设置为-7，一遍循环之后，数组就成为[-4,-3,-2,-7,8,2,-3,-1],然后第二次遍历，如果数字为正，就代表数组中不存在这个数，记录到答案中即可，这里要注意数组从0开始，而数字从1开始。</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></pre></td><td class="code"><pre><span class="line"> public static List<Integer> findDisappearedNumbers(int[] nums) {</span><br><span class="line"> for(int i=0;i<nums.length;i++){</span><br><span class="line"> int temp = Math.abs(nums[i]);</span><br><span class="line"> nums[temp-1]=-Math.abs(nums[temp-1]);</span><br><span class="line"> }</span><br><span class="line"> List<Integer>ans = new ArrayList(nums.length);</span><br><span class="line"> for(int i=0;i<nums.length;i++)</span><br><span class="line"> {</span><br><span class="line"> if(nums[i]>0){</span><br><span class="line"> ans.add(i+1);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return ans;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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