69. Sqrt(x) (Easy)

https://leetcode.com/problems/sqrtx/

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

Solutions

class Solution {
    public int mySqrt(int x) {
        if (x == 1) {
            return 1;
        }

        long least = 0;
        long most = x;

        while (least <= most) {
            long mid = (most * 1l + least * 1l) / 2;
            long squire = mid * mid;

            if (squire == x) {
                return (int) mid;
            } else if (squire > x) {
                most = (int) mid - 1;
            } else {
                // if mid is the largest integer of which the squire is the closest to but less than x
                least = (int) (mid + 1);
            }
        }

        // to reach here, least > most and most^2 < x < least^2
        return (int) most;
    }
}

Incorrect Solutions

References

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