https://leetcode.com/problems/valid-perfect-square/
https://www.hrwhisper.me/leetcode-valid-perfect-square/
http://blog.csdn.net/u014688145/article/details/69388673
需要注意一点！为了防止大数相乘溢出，这里判断语句为
public boolean isPerfectSquare(int num) { if (num == 0) return false; int end = binarySearch(num, num); return end == -1 ? false : true; } private int binarySearch(int num, int target){ int lf = 1, rt = num; //闭区间 while (lf < rt){ int mid = lf + (rt - lf) / 2; //防止溢出 if (target / mid > mid){ lf = mid + 1; }else{ rt = mid; } } if (target % rt == 0 && target / rt == rt) return rt; return -1; }
http://blog.csdn.net/qq508618087/article/details/51762492
这道题给了我们一个数，让我们判断其是否为完全平方数，那么显而易见的是，肯定不能使用brute force，这样太不高效了，那么最小是能以指数的速度来缩小范围，那么我最先想出的方法是这样的，比如一个数字49，我们先对其除以2，得到24，发现24的平方大于49，那么再对24除以2，得到12，发现12的平方还是大于49，再对12除以2，得到6，发现6的平方小于49，于是遍历6到12中的所有数，看有没有平方等于49的，有就返回true，没有就返回false
https://leetcode.com/discuss/110659/o-1-time-c-solution-inspired-by-q_rsqrt
Given a positive integer num, write a function which returns True if num is a perfect square else False.
Note: Do not use any built-in library function such as
sqrt.
Example 1:
Input: 16 Returns: True
Example 2:
Input: 14 Returns: FalseX. Binary Search
https://www.hrwhisper.me/leetcode-valid-perfect-square/
题意：给定一个数n，要求不使用sqrt函数判断该数是否为完全平方数
思路：二分。。 L = 1 , R = num / 2 +1开始枚举即可。。。
public boolean isPerfectSquare(int num) {
long L = 1, R = (num >> 1) + 1;
while (L <= R) {
long m = L + ((R - L) >> 1);
long mul = m * m;
if (mul == num) return true;
else if (mul > num) R = m - 1;
else L = m + 1;
}
return false;
}
需要注意一点！为了防止大数相乘溢出，这里判断语句为
target / mid > mid，把这部分解给排除后，循环外还需要增加一个额外的判断，如 42 / 6 > 6，但是它并不是Perfect Square.public boolean isPerfectSquare(int num) { if (num == 0) return false; int end = binarySearch(num, num); return end == -1 ? false : true; } private int binarySearch(int num, int target){ int lf = 1, rt = num; //闭区间 while (lf < rt){ int mid = lf + (rt - lf) / 2; //防止溢出 if (target / mid > mid){ lf = mid + 1; }else{ rt = mid; } } if (target % rt == 0 && target / rt == rt) return rt; return -1; }
http://blog.csdn.net/qq508618087/article/details/51762492
bool isPerfectSquare(int num) {
if(num < 1) return false;
if(num == 1) return true;
int left = 0, right = num/2;
while(left <= right)
{
long mid = (left+right)/2;
long val = mid*mid;
if(val == num) return true;
else if(val > num) right = mid-1;
else left = mid+1;
}
return false;
}
bool isPerfectSquare(int num) { long left = 0, right = num; while (left <= right) { long mid = left + (right - left) / 2, t = mid * mid; if (t == num) return true; else if (t < num) left = mid + 1; else right = mid - 1; } return false; }
X. Newton's method
long r = x;
while (r*r > x)
r = (r + x/r) / 2;
return r*r == x;
Apparently, using only integer division for the Newton method works. And I guessed that if I start at x, the root candidate will decrease monotonically and never get too small.
int mySqrt(int x) {
long long r = x;
while (r*r > x)
r = (r + x/r) / 2;
return r;
}
def isPerfectSquare(self, num):
"""
:type num: int
:rtype: bool
"""
x = num
while x * x > num:
x = (x + num / x) / 2
return x * x == num
This is a math problem：
1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
so 1+3+...+(2n-1) = (2n-1 + 1)n/2 = nn
1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
so 1+3+...+(2n-1) = (2n-1 + 1)n/2 = nn
public boolean isPerfectSquare(int num) {
int i = 1;
while (num > 0) {
num -= i;
i += 2;
}
return num == 0;
}
Let x be the number of iterations needed to solve the problem, and let Σ be the sum from i = 1 to x. Σ(1 + 2i) = n => x + 2Σi = n => x + 2(x(x+1)) = n => 2x^2 + 3x = n => x = [-3 +/- sqrt(9 + 8n)]/4 => you can see that n is in a square root term, so the complexity should be O(sqrt(n)).
bool isPerfectSquare(int num) { if (num == 1) return true; long x = num / 2, t = x * x; while (t > num) { x /= 2; t = x * x; } for (int i = x; i <= 2 * x; ++i) { if (i * i == num) return true; } return false; }
下面这种方法也比较高效，从1搜索到sqrt(num)，看有没有平方正好等于num的数：
bool isPerfectSquare(int num) { for (int i = 1; i <= num / i; ++i) { if (i * i == num) return true; } return false; }
https://leetcode.com/discuss/110659/o-1-time-c-solution-inspired-by-q_rsqrt
bool isPerfectSquare(int num) {
if (num < 0) return false;
int root = floorSqrt(num);
return root * root == num;
}
int32_t floorSqrt(int32_t x) {
double y=x; int64_t i=0x5fe6eb50c7b537a9;
y = *(double*)&(i = i-(*(int64_t*)&y)/2);
y = y * (3 - x * y * y) * 0.5;
y = y * (3 - x * y * y) * 0.5;
i = x * y + 1; return i - (i * i > x);
}
From https://en.wikipedia.org/wiki/Fastinversesquare_root
The following code is the fast inverse square root implementation from Quake III Arena, stripped of C preprocessor directives, but including the exact original comment text:
...
i = 0x5f3759df - ( i >> 1 ); // what the fuck?
...