LeetCode-Sqrt(x)

问题:实现  int sqrt(int x).

这里给出两种实现方法:一是二分搜索,二是牛顿迭代法。

  • 二分搜索
 1 int sqrt(int x) {
 2     long long i = 0;
 3     long long j = x / 2 + 1;
 4     while (i <= j)
 5     {
 6         long long mid = (i + j) / 2;
 7         long long sq = mid * mid;
 8         if (sq == x) return mid;
 9         else if (sq < x) i = mid + 1;
10         else j = mid - 1;
11     }
12     return j;
13 }
  • 牛顿迭代法
 1 int sqrt(int x) {
 2     if (x == 0) return 0;
 3     double last = 0;
 4     double res = 1;
 5     while (res != last)
 6     {
 7         last = res;
 8         res = (res + x / res) / 2;
 9     }
10     return int(res);
11 }
  • 此外,如果返回值不是整型
 1 double sqrt(double x) {
 2     if (x == 0) return 0;
 3     double last = 0.0;
 4     double res = 1.0;
 5     while (res != last)
 6     {
 7         last = res;
 8         res = (res + x / res) / 2;
 9     }
10     return res;
11 }

参考:http://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html

原文地址:https://www.cnblogs.com/MazeHong/p/7337898.html