[Leetcode 85] 69 Sqrt(x)
Problem:
Implement int sqrt(int x)
.
Compute and return the square root of x.
Analysis:
If we simply use the naive algorithm to go through all the numbers from 0 until the square root, it will exceed the time limit. The hero is binary search.
We want such a number between 0 and n inclusively that x*x <= n and (x+1)*(x+1) > n. Not exactly x*x = n because its the integer square root. With this exit condition, we can easily use binary search to get the final result.
Code:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 int sqrt(int x) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int s = 0, e = x; 7 8 while (s <= e) { 9 long long mid = (s + e) / 2, sq = mid * mid; 10 11 if (sq <= x && (mid+1) * (mid+1) > x) 12 return (int) mid; 13 else if (sq > x) 14 e = mid - 1; 15 else 16 s = mid + 1; 17 } 18 19 return -1; 20 } 21 };