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Sqrtx.java
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Sqrtx.java
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// Link: https://leetcode.com/problems/sqrtx/
// Difficulty: Easy
/*
* 72ms, 5%
*
* Keep doing x/i until x/i is smaller than i, then i-1 will be the result.
*/
class Solution {
public int mySqrt(int x) {
int i = 0;
for (; x / (i + 1) >= i + 1; ++i);
return i;
}
}
/*
* Time complexity: O(sqrt(x))
*
* Space complexity: O(1)
*
* NOTES:
* Here we use i+1 to handle the case where x=0
*
*/
/*
* Method-2: binary search
* 1ms, 99.99%
*
* To calculate sqrt(x), we need to find a number num, x/num >= num && x/(num+1) < num+1. E.g., x=8,
* we know that 8/2=4, 8/3=2, so sqrt(8) is 2.
*/
class Solution {
public int mySqrt(int x) {
if (x == 0) {
return 0;
}
int left = 1;
int right = x;
while (left <= right) {
int mid = left + (right - left) / 2;
if (x / mid >= mid && x / (mid + 1) < mid + 1) {
return mid;
} else if (x / mid < mid) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}
}
/*
* Time complexity: O(logx) = O(32) = O(1)
*
* Space complexity: O(1)
*
* NOTES:
*
*/