Question

以下算法如何计算完美正方形。我正在尝试使用这条逻辑来计算平方根。但我不明白它是如何计算平方根的。是否有任何算法可以做逻辑用这段代码写的？

public static boolean isPerfectSquare(BigDecimal num) {
            BigDecimal squareRoot = one;
            BigDecimal square = one;
            BigDecimal i = one;
            BigDecimal newSquareRoot;
            int comparison = -1;

            while (comparison != 0) {
                if (comparison < 0) {
                    i = i.multiply(two);
                    newSquareRoot = squareRoot.add(i).setScale(0, RoundingMode.HALF_UP);
                } else {
                    i = i.divide(two);
                    newSquareRoot = squareRoot.subtract(i).setScale(0, RoundingMode.HALF_UP);
                }

                if (newSquareRoot.compareTo(squareRoot) == 0) {
                    return false;
                }

                squareRoot = newSquareRoot;
                square = squareRoot.multiply(squareRoot);
                comparison = square.compareTo(num);
            }

            return true;
        }

Answer 1

我建议您添加一些System.out.println()来电，以了解自己的进展情况。

下面是我在完成相同操作并在101上运行后收到的输出。它只是增加猜测直到它太高，然后改进它的猜测，直到它找到完全匹配或确定它不能。

它的改进过程是逐渐减小步骤，直到它太低。然后它跳回来（步骤加倍）并且它再次开始向下踩。如果它到达其步长值小于1的点，则它放弃，因为参数不是完美的正方形。如果在任何步骤中，猜测的平方与参数匹配，那么您已经找到了平方根，因此您知道参数是一个完美的正方形。

   1 is lower  than 101: adding    2. New guess at square root is   3 (  9)
   9 is lower  than 101: adding    4. New guess at square root is   7 ( 49)
  49 is lower  than 101: adding    8. New guess at square root is  15 (225)
 225 is higher than 101: subbing   4. New guess at square root is  11 (121)
 121 is higher than 101: subbing   2. New guess at square root is   9 ( 81)
  81 is lower  than 101: adding    4. New guess at square root is  13 (169)
 169 is higher than 101: subbing   2. New guess at square root is  11 (121)
 121 is higher than 101: subbing   1. New guess at square root is  10 (100)
 100 is lower  than 101: adding    2. New guess at square root is  12 (144)
 144 is higher than 101: subbing   1. New guess at square root is  11 (121)
 121 is higher than 101: subbing 0.5. New guess at square root is  11 (121)
101 is not a perfect square

检查一个数字是否完美正方形

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