2021-03-20 09:16:07 +00:00
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> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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#### [35. 搜索插入位置](https://leetcode-cn.com/problems/search-insert-position/)
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2021-03-18 04:47:47 +00:00
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#### 题目描述
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> 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。
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>
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> 你可以假设数组中无重复元素。
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示例 1:
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> 输入: [1,3,5,6], 5
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> 输出: 2
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示例 2:
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> 输入: [1,3,5,6], 2
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> 输出: 1
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示例 3:
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> 输入: [1,3,5,6], 7
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> 输出: 4
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示例 4:
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> 输入: [1,3,5,6], 0
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> 输出: 0
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#### 题目解析
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这个题目完全就和咱们的二分查找一样，只不过有了一点改写，那就是将咱们的返回值改成了 left，具体实现过程见下图
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#### 题目代码
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```java
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class Solution {
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public int searchInsert(int[] nums, int target) {
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int left = 0, right = nums.length-1;
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//注意循环条件
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while (left <= right) {
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//求mid
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int mid = left + ((right - left ) >> 1);
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//查询成功
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if (target == nums[mid]) {
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return mid;
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//右区间
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} else if (nums[mid] < target) {
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left = mid + 1;
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//左区间
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} else if (nums[mid] > target) {
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right = mid - 1;
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}
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}
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//返回插入位置
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return left;
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}
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}
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```
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