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

Solution 1: Binary Search

We define the left boundary of the binary search as $left=0$, and the right boundary as $right=n-1$.

In each iteration, we calculate the middle position $mid=(left+right)/2$, and then compare the size of $nums[mid]$ and $target$:

• If $nums[mid] \geq target$, it means that $target$ is in the interval $[left, mid]$, so we update $right$ to $mid$;
• Otherwise, $target$ is in the interval $[mid+1, right]$, so we update $left$ to $mid+1$.

When $left \geq right$, we check if $nums[left]$ equals $target$. If it does, we return $left$, otherwise, we return $-1$.

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

Python3JavaC++GoRustJavaScriptC#

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

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

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

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

use std::cmp::Ordering;

impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let mut l = 0;
        let mut r = nums.len();
        while l < r {
            let mid = (l + r) >> 1;
            match nums[mid].cmp(&target) {
                Ordering::Less => {
                    l = mid + 1;
                }
                Ordering::Greater => {
                    r = mid;
                }
                Ordering::Equal => {
                    return mid as i32;
                }
            }
        }
        -1
    }
}

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
    let left = 0;
    let right = nums.length - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[mid] >= target) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return nums[left] == target ? left : -1;
};

public class Solution {
    public int Search(int[] nums, int target) {
        int left = 0, right = nums.Length - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[mid] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return nums[left] == target ? left : -1;
    }
}

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