leetcode 633. Sum of Square Numbers

Given a non-negative integer 

c

, your task
is to decide whether there're two integers 

a

 and 

b

 such
that a2 + b2 = c.
Example 1:

Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: 3
Output: False

这道题要注意两个测试用例：

2 , return true 。

2147482647，return false。

public boolean judgeSquareSum(int c) {
HashSet<Integer> set = new HashSet<Integer>();

for (int i = 0; i <= Math.sqrt(c); i++) {
set.add(i * i);
if (set.contains(c - i * i)) {
return true;
}
}
return false;
}

有大神用了类似于二分查找的思想。

public class Solution {
public boolean judgeSquareSum(int c) {
if (c < 0) {
return false;
}
int left = 0, right = (int)Math.sqrt(c);
while (left <= right) {
int cur = left * left + right * right;
if (cur < c) {
left++;
} else if (cur > c) {
right--;
} else {
return true;
}
}
return false;
}
}

这道题有solutions：https://leetcode.com/problems/sum-of-square-numbers/solution/

Approach #3 Using sqrt function [Accepted]

Algorithm

Check if \sqrt{c
- a^2}√​c−a​2​​​​​ turns
out to be an integer.

Java

public class Solution {
public boolean judgeSquareSum(int c) {
for (long a = 0; a * a <= c; a++) {
double b = Math.sqrt(c - a * a);
if (b == (int) b)
return true;
}
return false;
}
}

Complexity Analysis

Time complexity : O\big(\sqrt{c}log(c)\big)O(√​c​​​log(c)).
We iterate over \sqrt{c}√​c​​​ values
for choosing aa.
For every aa chosen,
finding square root of c
- a^2c−a​2​​ takes O\big(log(c)\big)O(log(c)) time
in the worst case.

Space complexity : O(1)O(1).
Constant extra space is used.

Approach #4 Using Binary Search [Accepted]

Algorithm

Another method to check if c
- a^2c−a​2​​ is
a perfect square, is by making use of Binary Search. 

Java

public class Solution {
public boolean judgeSquareSum(int c) {
for (long a = 0; a * a <= c; a++) {
int b = c - (int)(a * a);
if (binary_search(0, b, b))
return true;
}
return false;
}
public boolean binary_search(long s, long e, int n) {
if (s > e)
return false;
long mid = s + (e - s) / 2;
if (mid * mid == n)
return true;
if (mid * mid > n)
return binary_search(s, mid - 1, n);
return binary_search(mid + 1, e, n);
}
}

Complexity Analysis

Time complexity : O\big(\sqrt{c}log(c)\big)O(√​c​​​log(c)).
Binary search taking O\big(log(c)\big)O(log(c)) in
the worst case is done for \sqrt{c}√​c​​​ values
of aa.

Space complexity : O(log(c))O(log(c)).
Binary Search will take O(log(c))O(log(c)) space.

Approach #5 Fermat Theorem [Accepted]:

Algorithm

这个方法基于费马的平方和定理:

任何正整数 nn 是两个平方数之和的充分必要条件是: nn 的任意满足 (4k+3)(4k+3)  的素数因子都满足出现了偶数次。

Interested reader can refer to this documentation.

Java

public class Solution {
public boolean judgeSquareSum(int c) {
for (int i = 2; i * i <= c; i++) {
int count = 0;
if (c % i == 0) {
while (c % i == 0) {
count++;
c /= i;
}
if (i % 4 == 3 && count % 2 != 0)
return false;
}
}
return c % 4 != 3;
}
}

Complexity Analysis

Time complexity : O\big(\sqrt{c}log(c)\big)O(√​c​​​log(c)).
We find the factors of cc and
their count using repeated division. We check for the factors in the range [0,
\sqrt{c}][0,√​c​​​].
The maximum number of times a factor can occur(repeated division can be done) is log(n)log(n)(considering
2 as the only factor, c=2^xc=2​x​​.
Thus, x=log(c)x=log(c)).

Space complexity : O(1)O(1).
Constant space is used.