[leetcode] sqrt(int num)
Implement int sqrt(int x).
Compute and return the square root of x.
TestCases:
| input | output | expected | |
|---|---|---|---|
| 0 | 0 | 0 |
|
| 1 | 1 | 1 |
|
| 2 | 1 | 1 |
|
| 3 | 1 | 1 |
|
| 4 | 2 | 2 |
|
| 5 | 2 | 2 |
|
| 6 | 2 | 2 |
|
| 7 | 2 | 2 |
|
| 8 | 2 | 2 |
|
| 9 | 3 | 3 |
|
| 10 | 3 | 3 |
|
| 1024 | 32 | 32 |
|
| 8192 | 90 | 90 |
|
| 2147395599 | 46339 | 46339 |
|
| 2147395600 | 46340 | 46340 |
|
| 2147483647 | 46340 | 46340 |
|
要注意的问题:
1. sqrt(10)=3
2. int由32bit表示，不可以越界！一般思路：sqrt(x) < x/2, 从0-x/2开始做binary search. 但x>2^16时 (x/2)^2会int溢出。必须设定搜索上限
class Solution {
public:
int sqrt(int num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int start=0;
int end=0;
int tmp=num;
int digits=0;
//get the upperbound of its sqrt
while(tmp > 0)
{
digits++;
tmp=tmp/10;
}
for(int i=0;i<(digits+1)/2;i++)
end=end*10+9;
if(end > 46340) end=46340; //the largest for 32-bit integer in C++
if(end > num/2+1 )
end=(num+1)/2;
int med=0;
while(start <= end)
{
med=(start+end+1)/2;
if( med >= 46340)
return 46340;
if(med*med <= num && (med+1)*(med+1)>num)
return med;
if(med*med < num)
start=med;
else
end=med;
}
}
};
思路2： 牛顿搜索: http://en.wikipedia.org/wiki/Newton's_method#Square_root_of_a_number
class Solution {
public:
int sqrt(int x) {
float n_search_seed=10;
float num=(float)x;
float prev_seed=0;
float EPS = 0.00000001;
do //empiracle 20 loops should be good enough
{
prev_seed = n_search_seed;
n_search_seed=n_search_seed - (n_search_seed*n_search_seed - num)/(2*n_search_seed);
}while(abs(prev_seed - n_search_seed) > EPS);
int result=n_search_seed;
if(result*result > x)
result--;
return result;
}
};