LeetCode Python/Java/C++/JS > Array > 704. Binary Search > Solved in Java, Python, C++, JavaScript, C#, Go, Ruby > LeetCode GitHub Code or Repost
LeetCode link: 704. Binary Search, difficulty: Easy.
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 <= 10000104 < nums[i], target < 10000- All the integers in
numsare unique. numsis sorted in ascending order.
Intuition
Because it is an already sorted array, by using the middle value for comparison, half of the numbers can be eliminated each time.
Step by Step Solutions
The fastest and easiest way is to use the three indices left, right, and middle.
If nums[middle] > target, then right = middle - 1, otherwise, left = middle + 1.
Complexity
Time complexity
O(log N)
Space complexity
O(1)
Java #
class Solution {
public int search(int[] nums, int target) {
var left = 0;
var right = nums.length - 1;
while (left <= right) {
var middle = (left + right) / 2;
if (nums[middle] == target) {
return middle;
}
if (nums[middle] > target) {
right = middle - 1;
} else {
left = middle + 1;
}
}
return -1;
}
}
Python #
class Solution:
def search(self, nums: List[int], target: int) -> int:
left = 0
right = len(nums) - 1
while left <= right:
middle = (left + right) // 2
if nums[middle] == target:
return middle
if nums[middle] > target:
right = middle - 1
else:
left = middle + 1
return -1
C++ #
class Solution {
public:
int search(vector<int>& nums, int target) {
auto left = 0;
int right = nums.size() - 1; // Should not use 'auto' here because 'auto' will make this variable become `unsigned long` which has no `-1`.
while (left <= right) {
auto middle = (left + right) / 2;
if (nums[middle] == target) {
return middle;
}
if (nums[middle] > target) {
right = middle - 1;
} else {
left = middle + 1;
}
}
return -1;
}
};
JavaScript #
var search = function (nums, target) {
let left = 0
let right = nums.length - 1
while (left <= right) {
const middle = Math.floor((left + right) / 2)
if (nums[middle] == target) {
return middle
}
if (nums[middle] > target) {
right = middle - 1
} else {
left = middle + 1
}
}
return -1
};
C# #
public class Solution
{
public int Search(int[] nums, int target)
{
int left = 0;
int right = nums.Length - 1;
while (left <= right)
{
int middle = (left + right) / 2;
if (nums[middle] == target)
{
return middle;
}
if (nums[middle] > target)
{
right = middle - 1;
}
else
{
left = middle + 1;
}
}
return -1;
}
}
Go #
func search(nums []int, target int) int {
left := 0
right := len(nums) - 1
for left <= right {
middle := (left + right) / 2
if nums[middle] == target {
return middle
}
if nums[middle] > target {
right = middle - 1
} else {
left = middle + 1
}
}
return -1
}
Ruby #
def search(nums, target)
left = 0
right = nums.size - 1
while left <= right
middle = (left + right) / 2
return middle if nums[middle] == target
if nums[middle] > target
right = middle - 1
else
left = middle + 1
end
end
-1
end