LeetCode 069 Sqrt(x)
2016-07-08 23:06
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题目要求实现一个
由于输入输出都是int,最简单的方法就是从1开始递增枚举,如果平方大于x就停下来。但是太耗时了。我们可以用二分法枚举,实现上就是从大到小枚举每个二进制位是否为1。先找到满足
代码:
int sqrt(int x)。
由于输入输出都是int,最简单的方法就是从1开始递增枚举,如果平方大于x就停下来。但是太耗时了。我们可以用二分法枚举,实现上就是从大到小枚举每个二进制位是否为1。先找到满足
(2^n)^2<=x的最大n。然后从最高位到最低位判断
2^n中的n-1个二进制位是否能为1。
代码:
int mySqrt(int x) { if(x <= 0) return 0; int d = 1, i = 0; while(d * d <= x && d * d / d == d) d <<= 1, i++; i--; d = 1 << i; while(i--) { int k = d + (1 << i); if(k * k <= x && k * k / k == k) d += 1 << i; } return d; }
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