704.Binary Search
===

###### tags: `Easy`,`Array`,`Binary Search`

[704. Binary Search](https://leetcode.com/problems/binary-search/)

### 題目描述

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.

### 範例

**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` <= 10^4^
* -10^4^ < `nums[i]`, `target` < 10^4^
* All the integers in `nums` are **unique**.
* `nums` is sorted in ascending order.

### 解答

#### Javascript

function search(nums, target) {
  let min = 0;
  let max = nums.length - 1;

  while (max >= min) {
    const mid = Math.floor((max + min) / 2);
    if (nums[mid] === target) return mid;
    
    if (nums[mid] > target) {
      max = mid - 1;
    } else {
      min = mid + 1;
    }
  }

  return -1;
}

> [name=Marsgoat][time=Apr 1, 2023]

#### Python

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        L, R = 0, len(nums)
        while L < R:
            M = (L + R) // 2
            n = nums[M]
            if n == target:
                return M
            if n > target:
                R = M
            else:
                L = M + 1
        return -1

### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)