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704. Binary Search

Description

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.

Solutions

We define the left boundary $l=0$ and the right boundary $r=n-1$ for binary search.

In each iteration, we calculate the middle position $\textit{mid}=(l+r)/2$, then compare the size of $\textit{nums}[\textit{mid}]$ and $\textit{target}$.

  • If $\textit{nums}[\textit{mid}] \geq \textit{target}$, it means $\textit{target}$ is in the left half, so we move the right boundary $r$ to $\textit{mid}$;
  • Otherwise, it means $\textit{target}$ is in the right half, so we move the left boundary $l$ to $\textit{mid}+1$.

The loop ends when $l<r$, at this point $\textit{nums}[l]$ is the target value we are looking for. If $\textit{nums}[l]=\textit{target}$, return $l$; otherwise, return $-1$.

The time complexity is $O(\log n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

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class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums) - 1
        while l < r:
            mid = (l + r) >> 1
            if nums[mid] >= target:
                r = mid
            else:
                l = mid + 1
        return l if nums[l] == target else -1

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class Solution {
    public int search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return nums[l] == target ? l : -1;
    }
}

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class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return nums[l] == target ? l : -1;
    }
};

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func search(nums []int, target int) int {
    l, r := 0, len(nums)-1
    for l < r {
        mid := (l + r) >> 1
        if nums[mid] >= target {
            r = mid
        } else {
            l = mid + 1
        }
    }
    if nums[l] == target {
        return l
    }
    return -1
}

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function search(nums: number[], target: number): number {
    let [l, r] = [0, nums.length - 1];
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] >= target) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return nums[l] === target ? l : -1;
}

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