Skip to content

Latest commit

 

History

History
79 lines (59 loc) · 1.5 KB

README.md

File metadata and controls

79 lines (59 loc) · 1.5 KB

Description

Implement int sqrt(int x).

Compute and return the square root of x.

x is guaranteed to be a non-negative integer.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated.

Tags: Binary Search, Math

思路

题意是求平方根,参考 牛顿迭代法求平方根,然后再参考维基百科的 Integer square root 即可。

java:

class Solution {
    public int mySqrt(int x) {
        long n = x;
        while (n * n > x) {
            n = (n + x / n) >> 1;
        }
        return (int) n;
    }
}

kotlin(192ms/78.57%):

class Solution {
    fun mySqrt(x: Int): Int {
        var k = x.toLong()
        while (k * k > x)
            k = (k + x / k) / 2
        return k.toInt()
    }
}

javascript:

var mySqrt = function(x) {
    for(var i = 0; i <= x; i++) {
        if(i*i === x) {
            return i
        }
        if(i*i > x) {
			return i-1
        }
    }
};

结语

如果你同我们一样热爱数据结构、算法、LeetCode,可以关注我们 GitHub 上的 LeetCode 题解:LeetCode-Solution