假设500个小孩手拉着手围着一圈，数三就退出圈子，最后留在圈内的小孩是第几号？
有一种是用很传统的算法,用数组来编。
程序代码如下：
public class CountThreeDel1 { public static void main(String args[]) { boolean[] arr = new boolean[500]; for(int i=0;i<arr.length;i++){ arr[i] = true; } int leftCount = arr.length; int countMum = 0; int index = 0; while(leftCount>1) { if(arr[index]==true) { countMum ++; if(countMum == 3) { countMum = 0; arr[index] = false; leftCount --; } } index ++; if(index==arr.length) { index = 0; } } for(int i=0;i<arr.length;i++) { if(arr[i]==true) { System.out.println(i); int d = i + 1; System.out.println("所以剩下的是第" + d + "号!"); } } }}
采用面向对象的思路：
面向对象的思路，首先考虑这个问题域中有那些类，类里面有哪些属性，哪些方法
题中有一个小孩和由一个小孩围成的圈。
程序代码如下：
public class CountThreeDel2 { public static void main(String[] args) { KidCircle kc = new KidCircle(500); int countNum = 0; Kid k = kc.first; while(kc.count > 1) { countNum ++; if(countNum == 3){ countNum = 0; kc.delet(k); } k = k.right; } System.out.print("剩下的小孩编号是："); System.out.println(kc.first.id+1); } }class Kid { int id; Kid left; Kid right;}class KidCircle{ int count = 0; Kid first; Kid last; KidCircle(int n ) { for(int i=0;i<n;i++) { add(); } } void add() { Kid k = new Kid(); k.id = count; if(count <= 0) { first = k; last = k; k.left = k; k.right = k; } else { last.right = k; k.left = last; k.right = first; first.left = k; last = k; } count ++; } void delet(Kid k) { if(count <= 0) { return; } else if(count == 1) { first = last = null; } else { k.left.right = k.right; k.right.left = k.left; if (k == first) { first = k.right; } else if(k ==last) { last = k.left; } } count --; }}
