我在Java中玩了一些練習(xí)問題。我為下面給出的程序編寫了一個(gè)遞歸程序。我的解決方案是正確的，除了掛起（我相信）恢復(fù)到活動(dòng)狀態(tài)并更改遞歸方法的值。我還在調(diào)試模式下添加了 Eclipse 的屏幕截圖，其中顯示了線程堆棧。package com.nix.tryout.tests;/** * For given two numbers A and B such that 2 <= A <= B, * Find most number of sqrt operations for a given number such that square root of result is a whole number and it is again square rooted until either the  * number is less than two or has decimals.  * example if A = 6000 and B = 7000, sqrt of 6061 = 81, sqrt of 81 = 9 and sqrt of 9 = 3. Hence, answer is 3 *  * @author nitinramachandran * */public class TestTwo {    public int solution(int A, int B) {        int count = 0;        for(int i = B; i > A ; --i) {            int tempCount = getSqrtCount(Double.valueOf(i), 0);            if(tempCount > count) {                count = tempCount;             }        }        return count;    }    // Recursively gets count of square roots where the number is whole    private int getSqrtCount(Double value, int count) {        final Double sqrt = Math.sqrt(value);        if((sqrt > 2) && (sqrt % 1 == 0)) {            ++count;            getSqrtCount(sqrt, count);        }        return count;    }    public static void main(String[] args) {        TestTwo t2 = new TestTwo();        System.out.println(t2.solution(6550, 6570));    }} 上面的屏幕截圖來自我的調(diào)試器，我已經(jīng)圈出了線程堆棧。任何人都可以嘗試運(yùn)行該程序，并讓我知道問題是什么以及解決方案是什么？我可以想出一個(gè)非遞歸解決方案。