# 704. Binary-Search

difficulty: Easy

&#x20;section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }&#x20;

Given a **sorted** (in ascending order) integer array `nums` of `n` elements and a `target` value, write a function to search `target` in `nums`. If `target` exists, then return its index, otherwise return `-1`.

\
&#x20;**Example 1:**

```
Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4
```

**Example 2:**

```
Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1
```

**Note:**

1. You may assume that all elements in `nums` are unique.
2. `n` will be in the range `[1, 10000]`.
3. The value of each element in `nums` will be in the range `[-9999, 9999]`.

## Method One

```java
class Solution {
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        while(left <= right) {
            int pivot = left + (right - left ) / 2;
            if(target < nums[pivot]) {
                right = pivot - 1;
            }else if( target > nums[pivot]) {
                left = pivot + 1;
            }else {
                return pivot;
            }
        }
        return -1;
    }
}
​
```
