0069._Sqr(x).md

69. Sqrt(x)

难度:Easy

刷题内容

原题连接

• https://leetcode.com/problems/sqrtx/

内容描述

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:

Input: 4
Output: 2
Example 2:

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

思路1 - 时间复杂度: O(n)- 空间复杂度: O(1)******

遍历从0到x的所有数，计算这些数的平方，直到大于x。

class Solution {
public:
    int mySqrt(int x) {
        int i = 0;
        if(!x || x == 1)
            return x;
        for(;i < x;++i)
             if((long long)i * i > x)
                 break;
        return i - 1;
    }
};

思路2 - 时间复杂度: O(lgn)- 空间复杂度: O(1)******

很明显，上述方法我可以进一步优化，用二分法解决，以0为下界，x为上界，就可以进行二分搜索

class Solution {
public:
    int mySqrt(int x) {
        int l = 0, r= x;
        while(l < r)
        {
            long long mid = (l + r) / 2,temp = mid * mid;
            if(temp == x)
                return mid;
            if(temp < x)
                l = mid + 1;
            else
                r = mid - 1;
        }
        return l * l <= x ? l : l - 1;
    }
};