[leetcode] 69. Sqrt(x)

Description

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.

分析

题目的意思是：求一个数的平方根的整数部分。

1. 第一种：需要考虑两数相乘溢出的情况。主要思想是用了二分法。
2. 第二种方式就是瞎找了，当然要考虑数据溢出的情况。

代码一

class Solution {
public:
    int sqrt(int x) {
        if(x<2){
            return x;
        }
        int low=1;
        int high=x/2;
        int mid=0;
        while(low<=high){
            mid=(high+low)/2;
            if(x/mid>mid){
                low=mid+1;
            }else if(x/mid==mid){ //不用x > mid * mid 会溢出
                return mid;
            }
            else{
                high=mid-1;
            }
        }
        if(x/mid<mid){
            return mid-1;
        }
        return mid;
    }
};

代码二

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

参考文献

​​[编程题]sqrtx​​​​[LeetCode] Sqrt(x) 求平方根​​