LeetCode 69. Sqrt(x)

分析

难度 易

来源

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

题目

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4

Output: 2

Example 2:

Input: 8

Output: 2

Explanation: The square root of 8 is 2.82842..., and since 

             the decimal part is truncated, 2 is returned.

解答

方法1

package LeetCode;
/*
蛮力，Runtime: 63 ms, faster than 6.02% of Java online submissions for Sqrt(x).
 */
public class L69_SqrtX {
    public int mySqrt(int x) {
        int res=0;
        int curSquare=0;//记录上一轮*方数，
        int nextSquare=0;
        for(int i=0;i<=x/2;i++)
        {
            nextSquare=(i+1)*(i+1);
            if(nextSquare>x||nextSquare<curSquare)//如果*方溢出，一定是大于x的
                break;
            else{
                res++;
                curSquare=nextSquare;
            }
        }
        return res;
    }

    public static void main(String[] args){
        L69_SqrtX l69=new L69_SqrtX();
        System.out.println(l69.mySqrt(2147395600));
    }
}

 

方法2 牛顿法

public int mySqrt (int x) {
    if (x <= 1)
        return x;
    double assume = x / 2;
    while (Math.abs(Math.pow(assume, 2) - x) >=1) {
        assume = (assume + x/assume) / 2;//求当前值与除以x的结果的均值，故能不断接**方根
    }
    return (int) Math.floor(assume);
}

 

方法3 //二分查找

public int mySqrt (int x) {
    if (x <= 1)
        return x;
    int left=1,right=Integer.MAX_VALUE;
    int res=0;
    while(left<right){
        res=left+(right-left)/2;//防止溢出
        if(res>x/res){
            right=res;
        }else{
            left=res;
        }
        if(right-left<=1)
            break;
    }
    return (left+right)/2;
}