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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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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
        }
    }
}

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/**
 * @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;
};

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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;
    }
}

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