Problem Description
Given a non-negative integer x, compute and return the square root of x. Leetcode: Sqrt(x)
Constraints:
0 <= x <= 2^31 - 1
Solution
The important part in this code is to realize that m*m can lead to overflow. To handle this, we can do:
m == x/m
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class Solution {
public int mySqrt(int x) {
int l=1, h=x, root=0;
while(l<=h){
int m = (l+h)/2;
if(m == x/m) return m; //m*m can lead to overflow.
if(m>x/m){
h = m-1;
}else{
root = m;
l = m+1;
}
}
return root;
}
}