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69. Sqrt(x)

Description

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.

You must not use any built-in exponent function or operator.

  • For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

 

Example 1:

 Input: x = 4 Output: 2 Explanation: The square root of 4 is 2, so we return 2.

Example 2:

 Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.

 

Constraints:

  • 0 <= x <= 231 - 1

Solutions

Solution 1: Binary Search

We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = x$, then we search for the square root within the range $[l, r]$.

In each step of the search, we find the middle value $mid = (l + r + 1) / 2$. If $mid > x / mid$, it means the square root is within the range $[l, mid - 1]$, so we set $r = mid - 1$. Otherwise, it means the square root is within the range $[mid, r]$, so we set $l = mid$.

After the search ends, we return $l$.

The time complexity is $O(\log x)$, and the space complexity is $O(1)$.

  • class Solution { public int mySqrt(int x) { int l = 0, r = x; while (l < r) { int mid = (l + r + 1) >>> 1; if (mid > x / mid) { r = mid - 1; } else { l = mid; } } return l; } } 
  • class Solution { public: int mySqrt(int x) { int l = 0, r = x; while (l < r) { int mid = (l + r + 1ll) >> 1; if (mid > x / mid) { r = mid - 1; } else { l = mid; } } return l; } }; 
  • class Solution: def mySqrt(self, x: int) -> int: l, r = 0, x while l < r: mid = (l + r + 1) >> 1 if mid > x // mid: r = mid - 1 else: l = mid return l 
  • func mySqrt(x int) int { return sort.Search(x+1, func(i int) bool { return i*i > x }) - 1 } 
  • /** * @param {number} x * @return {number} */ var mySqrt = function (x) { let [l, r] = [0, x]; while (l < r) { const mid = (l + r + 1) >> 1; if (mid > x / mid) { r = mid - 1; } else { l = mid; } } return l; }; 
  • public class Solution { public int MySqrt(int x) { int l = 0, r = x; while (l < r) { int mid = (l + r + 1) >>> 1; if (mid > x / mid) { r = mid - 1; } else { l = mid; } } return l; } } 
  • impl Solution { pub fn my_sqrt(x: i32) -> i32 { let mut l = 0; let mut r = x; while l < r { let mid = (l + r + 1) / 2; if mid > x / mid { r = mid - 1; } else { l = mid; } } l } }

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