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.
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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;
}
};
˼·2 **- ʱ¼ä¸´ÔÓ¶È: O(lgn)*- ¿Õ¼ä¸´ÔÓ¶È: O(1)***
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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;
}
};
