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LeetCode 35. Search Insert Position.html
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<h2 id="35-Search-Insert-Position（二分查找）"><a href="#35-Search-Insert-Position（二分查找）" class="headerlink" title="35. Search Insert Position（二分查找）"></a>35. Search Insert Position（二分查找）</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/search-insert-position" target="_blank" rel="noopener">https://leetcode-cn.com/problems/search-insert-position</a></p>
<h3 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h3><p>给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。</p>
<p>你可以假设数组中无重复元素。</p>
<p>示例 1:</p>
<p> 输入: [1,3,5,6], 5<br> 输出: 2<br>示例 2:</p>
<p> 输入: [1,3,5,6], 2<br> 输出: 1<br>示例 3:</p>
<p> 输入: [1,3,5,6], 7<br> 输出: 4<br>示例 4:</p>
<p> 输入: [1,3,5,6], 0<br> 输出: 0</p>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>二分查找加了一些别的东西，升序数组，找到返回下标，未找到就返回可以加入的位置.<br>代码和正常的二分查找相同，需要注意的是，最后一个返回的应该是left，因为在最后一次循环中，left=right，然后left++，这个位置，就是应该存放的位置。</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></pre></td><td class="code"><pre><span class="line"> public static int searchInsert(int[] nums, int target) {</span><br><span class="line"> if (nums.length == 0) {</span><br><span class="line"> return 0;</span><br><span class="line"> }</span><br><span class="line"> int left = 0;</span><br><span class="line"> int right = nums.length - 1;</span><br><span class="line"></span><br><span class="line"> while (left <= right) {</span><br><span class="line"> int mid = left + (right - left) / 2;</span><br><span class="line"> if (nums[mid] == target) {</span><br><span class="line"> return mid;</span><br><span class="line"> } else if (nums[mid] < target) {</span><br><span class="line"> left = mid + 1;</span><br><span class="line"> } else if (nums[mid] > target) {</span><br><span class="line"> right = mid - 1;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return left;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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