704. Binary Search

Problem

Tags: Array, 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 <= 10^4
• -10^4 < nums[i], target < 10^4
• All the integers in nums are unique.
• nums is sorted in ascending order.

Code

C

// 704. Binary Search (7/22/54201)
// Runtime: 32 ms (91.95%) Memory: 7.22 MB (21.35%) 

int search (int nums[], int nums_size, int target) {
    int upper = nums_size - 1, lower = 0;
    
    while (lower <= upper) {
        int mid = (upper + lower) / 2;
        
        if (nums[mid] == target) {
            return mid;
        }
        
        if (nums[mid] > target) {
            upper = mid - 1;
        }
        else {
            lower = mid + 1;
        }
    }
    
    return -1;
}

JS

// 704. Binary Search (11/24/53712)
// Runtime: 60 ms (92.20%) Memory: 42.49 MB (93.36%) 

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
function search(nums, target) {
    let left = 0,
        right = nums.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (nums[mid] === target) return mid;
        if (left === right) break;
        if (nums[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}