mirror of
https://github.com/chefyuan/algorithm-base.git
synced 2024-12-01 08:18:55 +00:00
65 lines
1.4 KiB
Markdown
65 lines
1.4 KiB
Markdown
|
### leetcode35搜索插入位置
|
|||
|
|
|
|||
|
|
#### 题目描述
|
|||
|
|
|
|||
|
|
> 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。
|
|||
|
|
>
|
|||
|
|
> 你可以假设数组中无重复元素。
|
|||
|
|
|
|||
|
|
示例 1:
|
|||
|
|
|
|||
|
|
> 输入: [1,3,5,6], 5
|
|||
|
|
> 输出: 2
|
|||
|
|
|
|||
|
|
示例 2:
|
|||
|
|
|
|||
|
|
> 输入: [1,3,5,6], 2
|
|||
|
|
> 输出: 1
|
|||
|
|
|
|||
|
|
示例 3:
|
|||
|
|
|
|||
|
|
> 输入: [1,3,5,6], 7
|
|||
|
|
> 输出: 4
|
|||
|
|
|
|||
|
|
示例 4:
|
|||
|
|
|
|||
|
|
> 输入: [1,3,5,6], 0
|
|||
|
|
> 输出: 0
|
|||
|
|
|
|||
|
|
#### 题目解析
|
|||
|
|
|
|||
|
|
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
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
#### 题目代码
|
|||
|
|
|
|||
|
|
```java
|
|||
|
|
class Solution {
|
|||
|
|
public int searchInsert(int[] nums, int target) {
|
|||
|
|
|
|||
|
|
int left = 0, right = nums.length-1;
|
|||
|
|
//注意循环条件
|
|||
|
|
while (left <= right) {
|
|||
|
|
//求mid
|
|||
|
|
int mid = left + ((right - left ) >> 1);
|
|||
|
|
//查询成功
|
|||
|
|
if (target == nums[mid]) {
|
|||
|
|
return mid;
|
|||
|
|
//右区间
|
|||
|
|
} else if (nums[mid] < target) {
|
|||
|
|
left = mid + 1;
|
|||
|
|
//左区间
|
|||
|
|
} else if (nums[mid] > target) {
|
|||
|
|
right = mid - 1;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
//返回插入位置
|
|||
|
|
return left;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
|
|||
|
|
|