367. effective complete square dichotomy

367. Effective complete square number

Given a Positive integer num , Write a function , If num It's a complete square , Then return to true , Otherwise return to false .

Advanced ： Don't Use any built-in library functions , Such as sqrt .

Example 1：

Input ：num = 16
Output ：true

Example 2：

Input ：num = 14
Output ：false

We should first think of the dichotomy , It's important , Can greatly reduce the time complexity , We can learn more about this method ：

bool isPerfectSquare(int num){
    
    int low=0;
    int high=num;
    long long p=(low+high)/2;
       //printf("%d ",p);
    while(p*p!=num&&low<=high){
    
        if(p*p>num){
    
            high=p-1;
            p=(low+high)/2;
        }
        else{
    
            low=p+1;
              p=(low+high)/2;

        }
    }
    printf("%d ",p);
    if(p*p==num){
    
        return true;
    }
    return false;
    

}