---
# System prepended metadata

title: BinarySearch704
tags: [binarySearch, leetcode, easy]

---


# BinarySearch704

###### tags: `leetcode`,`binarySearch`,`easy`

**BinarySearch704**


>ref: https://leetcode.com/problems/binary-search/
>
Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

>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
 

>Constraints:
1 <= nums.length <= 104
-104 < nums[i], target < 104
All the integers in nums are unique.
nums is sorted in ascending order.



>1.timeResolution:O(log N)故需要對半拆
>2. looping過程中start 會一直推進,當start跟end差一時會檢驗的是start,收斂到 start==end時需進行最後檢驗

```java=
public int search(int[] nums, int target) {
		
		int len = nums.length;

		int start = 0;
		int end = len - 1;

		while (start <= end) {
			int med = start + (end - start) / 2;
                       //收斂時檢驗
			if (start == end && nums[med] != target) {
				break;
			}

			if (nums[med] > target) {
				end = med;
			} else if (nums[med] < target) {
				start = med + 1;
			} else {
				return med;
			}
		}

		return -1;

	}
```
    
    
    
