LeetCode之Sqrt(x)

1、题目

Implement int sqrt(int x).

Compute and return the square root of x.

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2、代码实现

public class Solution {public int mySqrt(int x) {if (x < 0)return -1;if (x == 0)return 0; if (x == 1) return 1;int max = x / 2 + 1;int min = 1;while (min <= max) {int mid = (min + max) / 2;if (mid <= x / mid && x / (mid + 1) < mid + 1)return mid;if (mid > x / mid) max = mid - 1;elsemin = mid + 1;}return 0;}
}

3、注意的地方

1）、第一要记得判断条件是

mid * mid <= x && (mid + 1) * (mid + 1) >x

2）、第二要防止数据超过整形数据范围，所以需要把条件转化为

mid <= x / mid && x / (mid + 1) < mid + 1

3）、逻辑要清晰，先求max,min, next we will get mid, and condition is mid * mid <= x && (mid + 1) * (mid + 1) > x, and by condition ,do not remember max = mid -1, min = mid +1 or max = mid, min = mid;

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LeetCode之Sqrt(x)

1、题目

2、代码实现

3、注意的地方

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