Leecode69 sqrtx

Title description

Implement the function int sqrt (int x).
Calculate and return the square root of x

analysis

• The sum of squares can be reduced to a sequence by the concept of sequence.
• Find the first number in the sequence that is greater than x, and the subscript can be used to find the result.

java code

public class Solution {
    public int sqrt(int x) {
        if(x <= 0){
            return 0;
        }
        //使用累加法 1 4 9 16 25 ****        
        //这些数的间隔是 3 5 7 9 ***
        //间隔自增2     
        int res = 0;
        int delta = 1;
        int index = 0;
        //找出第一个大于x的res和index
       while(res <= x){
           //如果x比较大 接近于 Integer.MAX_VALUE。
           //sqrt（Integer.MAX_VALUE）^2 可能比Integer.MAX_VALUE小很多
           //如果继续增量delta，会出现错误，而事实上，这些数x的解就已经算出来了
           if(Integer.MAX_VALUE - res < delta){
               return index ;
           }
            res += delta;
            delta += 2;
            index ++;
        }
        //出来循环的 一定是大于x的res
        return index - 1;
    }
}