LeetCode 69. Sqrt(x)

Description

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

my program

思路：看似很简单的问题，可以不断的优化算法，最开始容易想到的是定义一个num使其递增，直到num*num > x，可是未考虑num*num越界的问题，后更改判断条件为num <= x/num，可以消除越界问题。但是此算法超出此题要求的时间范围，即Time Limit Exceeded。

class Solution {
public:
    int mySqrt(int x) {
        if(0 == x) return 0;
        int num = 1;
        while(num <= x/num){
            num++;
        }
        if(num*num == x)
            return num;
        else
            return num-1;
    }
};

Submission Details
1017 / 1017 test cases passed.
Status: Time Limit Exceeded
Last executed input:
1579205274

二分法

时间复杂度为O(logN)，这样就不会Time Limit Exceeded

class Solution {  
public:  
    int sqrt(int x) {  
        double begin = 0;  
        double end = x;  
        double result = 1;  
        double mid = 1;  
        while(abs(result-x) > 0.000001){  
            mid = (begin+end)/2;  
            result = mid*mid;  
            if(result > x)    
                end = mid;  
            else begin = mid;  
        }  
        return (int)mid;  
    }  
};  

Submission Details
1017 / 1017 test cases passed.
Status: Accepted
Runtime: 6 ms

牛顿迭代法

牛顿迭代法（Newton’s method）又称为牛顿-拉夫逊（拉弗森）方法（Newton-Raphson method），它是牛顿在17世纪提出的一种在实数域和复数域上近似求解方程的方法。

class Solution {  
public:  
    int mySqrt(int x) {  
        long r = x;
        while (r*r > x)
            r = (r + x/r) / 2;
        return r;
    }  
};  

Submission Details
1017 / 1017 test cases passed.
Status: Accepted
Runtime: 6 ms