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