LEETCODE ALGORITHM:69. Sqrt(x)

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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:

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Input: 4
Output: 2

Example 2:

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Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

题解

二分法

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class Solution {
public:
    int mySqrt(int x) {
        //if(x==0) return 0;
        int left=0,right=x,ans=-1;
        while(left<=right)
        {
            int mid=(left+right)/2;
            if((long long)mid*mid>x) right=mid-1;
            else if(mid*mid<x)
            {
                ans=mid;
                left=mid+1;
            }
            else return mid;
        }
        return ans;
    }
};

牛顿迭代法

相当于求\(x^2-C=0\)的零点，从\(x=C\)开始找改点切线与x轴的交点，作为下一次起点

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class Solution {
public:
    int mySqrt(int x) {
        if (x == 0) {
            return 0;
        }

        double C = x, x0 = x;
        while (true) {
            double xi = 0.5 * (x0 + C / x0);
            if (fabs(x0 - xi) < 1e-7) {
                break;
            }
            x0 = xi;
        }
        return int(x0);
    }
};