窮舉法

package cn.itcast.recursion;    public class ShareWine {        public static void main(String[] args) {          ShareWine shareWine = new ShareWine();          shareWine.backBottle(12, 0, 0);      }        //定義三個瓶子的容量      private int b1 = 12;      private int b2 = 8;      private int b3 = 5;      private int m = 6;//目標酒量        //假設一開始，12,0,0      public void backBottle(int bb1, int bb2, int bb3) {          System.out.println("bb1酒量:" + bb1 + "   bb2酒量:" + bb2 + "   bb3酒量:" + bb3);          if (bb1 == m || bb2 == m || bb3 == m) {              System.out.println("find the bottle");              return;          }          //瓶子2還有酒&&瓶子3沒滿          if (bb2 != 0 && bb3 != b3) {              //沒到5升，倒不滿              if (bb2 + bb3 <= b3) {                  backBottle(bb1, 0, bb2 + bb3);              } else {                  //第二個參數：剛開始bb2有那麼多酒，倒進bb3之後，bb3滿了，bb2就等於之前的bb2 -（滿了之後的b3-沒倒之前的bb3）                  backBottle(bb1, bb2 - (b3 - bb3), b3);              }              //瓶子3滿了，往瓶子1倒          } else if (bb3 == b3) {              //沒到12升，倒不滿              if (bb3 + bb1 <= b1) {                  backBottle(bb1 + bb3, bb2, 0);              } else {                  backBottle(b1, bb2, bb3 - (b1 - bb1));              }              //從瓶子1往瓶子2倒酒          } else if (bb2 == 0) {              if (bb1 >= b2) {                  backBottle(bb1 - b2, b2, bb3);              } else {                  backBottle(0, bb1, bb3);              }          }      }  }