LeetCode之Sqrt(x)
【摘要】 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...
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
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public class Solution {
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public int mySqrt(int x) {
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if (x < 0)
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return -1;
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if (x == 0)
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return 0;
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if (x == 1)
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return 1;
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int max = x / 2 + 1;
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int min = 1;
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while (min <= max) {
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int mid = (min + max) / 2;
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if (mid <= x / mid && x / (mid + 1) < mid + 1)
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return mid;
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if (mid > x / mid)
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max = mid - 1;
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else
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min = mid + 1;
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}
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return 0;
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}
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}
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;
文章来源: chenyu.blog.csdn.net，作者：chen.yu，版权归原作者所有，如需转载，请联系作者。
原文链接：chenyu.blog.csdn.net/article/details/70042243
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