633. Sum of Square Numbers
Explanation:
To solve this problem, we can use a two-pointer approach. We initialize two pointers, left at 0 and right at the square root of c. At each step, we calculate the sum of the squares of the values at these pointers. If the sum is equal to c, we return true. If the sum is less than c, we increment the left pointer to increase the sum, and if the sum is greater than c, we decrement the right pointer to decrease the sum. We continue this process until left is less than or equal to right.
Time complexity: O(sqrt(c)) Space complexity: O(1)
:
class Solution {
public boolean judgeSquareSum(int c) {
int left = 0, right = (int)Math.sqrt(c);
while (left <= right) {
int sum = left*left + right*right;
if (sum == c) {
return true;
} else if (sum < c) {
left++;
} else {
right--;
}
}
return false;
}
}Code Editor (Testing phase)
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