# 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

• 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;
}
};

• 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


• 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
}

• /**
* @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;
};


• 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
}
}


• 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;
}
}

• 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;
}