# Learn algorithm together – 704 Binary search

Time：2022-1-16

### 1、 Title

LeetCode-704. Binary search

Difficulty: simple
Given an N-element ordered (ascending) integer array nums and a target value target, write a function to search the target in nums. If the target value exists, return the subscript, otherwise return – 1.

``````Example 1:
Input: num = [- 1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 appears in nums and the subscript is 4

Example 2:
Input: num = [- 1,0,3,5,9,12], target = 2
Output: - 1
Explanation: 2 does not exist in nums, so - 1 is returned

Tips:
You can assume that all elements in nums are not repeated.
N will be between [1, 10000].
Each element of num will be between [- 9999, 9999].``````

### 2、 Problem solving ideas

Binary search is an algorithm based on comparing the target value with the middle element of the array.

• If the target value is equal to the intermediate element, the target value is found.
• If the target value is small, continue to search on the left.
• If the target value is large, continue searching on the right.

### 3、 Implementation process

#### c++

``````class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0,right = nums.size() - 1,mid;

while(left <= right){
mid =(left+right)/2;
if(nums[mid] == target){
return mid;
}else if(nums[mid] > target){
right = mid - 1;
}else{
left = mid + 1;
}
}
return -1;
}
};
``````

#### PHP

``````class Solution {

/**
* @param Integer[] \$nums
* @param Integer \$target
* @return Integer
*/
function search(\$nums, \$target) {
\$left = 0;
\$right = count(\$nums) - 1;

while(\$left <= \$right){
\$mid =(int)((\$left+\$right)/2);
if(\$nums[\$mid] == \$target){
return \$mid;
}else if(\$nums[\$mid] > \$target){
\$right = \$mid - 1;
}else{
\$left = \$mid + 1;
}
}
return -1;
}
}
``````

#### JavaScript

``````/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var search = function(nums, target) {
let left = 0,right = nums.length - 1,mid;

while(left <= right){
mid = Math.floor((left+right)/2);
if(nums[mid] == target){
return mid;
}else if(nums[mid] > target){
right = mid - 1;
}else{
left = mid + 1;
}
}
return -1;
};
``````

### 4、 Summary

The time complexity of binary search is O (logn) and the space complexity is O (1).