README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0704.Binary%20Search/README_EN.md	Array Binary Search

704. Binary Search

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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

Solution 1: Binary Search

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 $\text{mid}=(l+r)/2$, then compare the size of $\text{nums}[\text{mid}]$ and $\text{target}$.

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

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

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let mut l: usize = 0;
        let mut r: usize = nums.len() - 1;
        while l < r {
            let mid = (l + r) >> 1;
            if nums[mid] >= target {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        if nums[l] == target {
            l as i32
        } else {
            -1
        }
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
    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;
};

C#

public 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;
    }
}