Leetcode : Sqrt

url :

https://leetcode.com/problems/sqrtx/description/

描述

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

思路二分法

代码

/** * Created by hadoop on 17-10-3. */public class MySqrt {    public int mySqrt(int x) {        return (int)sqrtHelper2(x);    }//非递归    private long sqrtHelper2(int x){        long start = 0;        long end = x;        long mid = 0;        long temp = 0;        long res = -1;        while(start <= end) {            mid = (start+end)/2;            temp = mid * mid;            if(temp == x) {                res = mid;                break;            }else if(temp > x){                end = mid-1;            }else {                start = mid+1;                res = mid;            }        }        return res;    }//递归    private long sqrtHelper(long start, long end, int x) {        if (start > end) return -1;        long mid = (start + end) / 2;        if (mid * mid == x) {//注意mid乘积之后数值的取值范围，这里使用long类型            return mid;        } else if (mid * mid < x) {            long temp = sqrtHelper(mid + 1, end, x);            if (temp == -1) {                return mid;            } else {                return temp;            }        } else {            long temp = sqrtHelper(0, mid-1 , x);            if(temp == -1){                return mid;            }else{                return temp;            }        }    }//测试    public static void main(String[]args){        MySqrt mySqrt = new MySqrt();        int x = mySqrt.mySqrt(2147395599);        System.out.println(x);    }}