Easy
O(n)的方法
class Solution {
public boolean isPerfectSquare(int num) {
if (num <= 0){
return false;
}
if (num == 1){
return true;
}
for(long i = 1; i*i <= num; i++){
if (i * i == num){
return true;
}
}
return false;
}
}
O(logn)的方法用的是Binary Search, 用的自己最熟的模版，注意越界所以用long.
class Solution {
public boolean isPerfectSquare(int num) {
if (num <= 0){
return false;
}
if (num == 1){
return true;
}
long start = 1;
long end = num;
while (start + 1 < end){
long mid = start + (end - start) / 2;
long square = mid * mid;
if (square < num){
start = mid;
} else if (square > num){
end = mid;
} else {
return true;
}
}
return false;
}
}
