LeetCode

Sqrt(x)

2013.12.22 04:02

Implement int sqrt(int x).

Compute and return the square root of x.

Solution 1:

　　The square root can be calculated using a binary iteration, with an initial interval of [1, x - 1].

　　Time complexity is O(log2(x)), space complexity is O(1).

Accepted code:

// 2WA, 1AC, binary search + long long int
class Solution {
public:
    int sqrt(int x) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        long long int xx = x;
        
        if(x <= 0){
            return 0;
        }else if(x < 4){
            return 1;
        }
        
        long long int ll, rr, mm;
        ll = 1;
        rr = xx - 1;
        while(rr - ll > 1){
            mm = (ll + rr) / 2;
            // 1WA here, wrong side of '<'
            if(xx < mm * mm){
                rr = mm;
            }else{
                ll = mm;
            }
        }
        
        // 1WA here, return $ll, not $rr
        return ll;
    }
};

Solution 2:

　　Actually there is no need to use long long int to avoid integer overflow, since "xx < mm * mm" can be replaced with "x / mm < mm".

　　Integer underflow is usually safer than overflow, especially when it converges to 0, that's why we use it for some convenience.

　　Time complexity is O(log2(x)), space complexity is O(1).

Accepted code:

// 1AC, integer overflow can be avoided by using '/' instead of '*'.
class Solution {
public:
    int sqrt(int x) {
        if(x <= 0){
            return 0;
        }else if(x < 4){
            return 1;
        }
        
        long long int ll, rr, mm;
        ll = 1;
        rr = x - 1;
        while(rr - ll > 1){
            mm = (ll + rr) / 2;
            if(x / mm < mm){
                rr = mm;
            }else{
                ll = mm;
            }
        }
        
        return ll;
    }
};