leetcode:69. Sqrt(x)

描述

Implement int sqrt(int x).

Compute and return the square root of x.

思路

使用二分法搜索。
这里我一开始犯了这么一个错误，mid*mid很容易超过计算机能够表示的阈值，这里只要把转换成mid与x/mid比较就好了

代码

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

结果

他山之玉

C++ O(n) solutioin

long r = x;    while (r*r > x)        r = (r + x/r) / 2;    return r;

Java O(n) solution

 public int mySqrt(int x) {        if (x < 1) return x;        int lo = 1, hi = x, mid = 0;        while (lo <= hi){            mid = (hi - lo) / 2 + lo;            if (mid > x / mid) hi = mid - 1;            else if (mid < x / mid) lo = mid + 1;            else return mid;        }        return hi;    }

python O(n) solution

class Solution(object):    def mySqrt(self, x):        return int(x**(1/float(2)))

这个思路应该算作弊吧，不过厉害了