LeetCode刷题笔记--69. Sqrt(x)

69. Sqrt(x)

Easy

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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.

用二分法来做，要注意1，0等特殊情况。二分法的边界很特别，要好好理解一下。另外if那里不能写成(M*M<=x)这样会报错超过范围。

class Solution {
public:
    int mySqrt(int x) {
        if(x<=0)return 0;   
        if(x==1)return 1;
        int L=0;
        int R=x;
        while(L<R)
        {
            int M=L+(R-L)/2;
            if(M<=x/M)L=M+1;
            else R=M;
        }
        return R-1;
    }
};