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