趣味算法——青蛙过河(JAVA)

　　青蛙过河是一个非常有趣的智力游戏，其大意如下： 一条河之间有若干个石块间隔，有两队青蛙在过河，每队有3只青蛙，这些青蛙只能向前移动，不能向后移动，且一次只能有一只青蛙向前移动。在移动过程中，青蛙可以向前面的空位中移动，不可以一次跳过两个位置，但是可以跳过对方一只青蛙进入到前面的一个空位。问两队青蛙该如何移动才能用最少的步数分别走向对岸？（ → → → □ ← ← ← ）可能3只青蛙太少了，心算也不难。如果有100只青蛙呢？

/**
* 青蛙过河
* @author rubekid
*
*/
public class RiverFrog {

public static final int LEFT_FROG = -1;

public static final int RIGHT_FROG = 1;

public static final int STONE = 0;

private int[] frogs;

private int zeroIndex;

private int length;

private int step = 0;

public RiverFrog(int number) {
frogs = new int[number * 2 +1];
length = frogs.length;
zeroIndex = length /2;
for(int i=0; i< number; i++){
frogs[i] = LEFT_FROG;
}
frogs[zeroIndex] = STONE;
for(int i=0; i< number; i++){
frogs[i+ zeroIndex + 1] = RIGHT_FROG;
}

}

public void run(){

while(!isMoveEnd(LEFT_FROG) || !isMoveEnd(RIGHT_FROG)){
int left = zeroIndex - 1;
int right = zeroIndex+1;

if(left>-1 && right <length){
if(frogs[left] != frogs[right]){
if(frogs[left] == LEFT_FROG){
if(left > 0 && frogs[left-1] == RIGHT_FROG){//若移动right，则在中间有两只RIGHT并排
this.move(right);
}
else{
this.move(left);
}
}
else if(left > 0 && frogs[left-1]==LEFT_FROG ){
this.move(left-1);
}
else if(right <= length && frogs[right+1] == RIGHT_FROG){
this.move(right+1);
}
}
else{
if(frogs[left] == RIGHT_FROG){
if(left > 0 && frogs[left-1] == LEFT_FROG){
this.move(left - 1);
}
else if(right+1 < length && frogs[right+1] == RIGHT_FROG){
this.move(right+1);
}
else if(frogs[right] == RIGHT_FROG){
this.move(right);
}
}
else if(frogs[right] == LEFT_FROG){
if(right+1 < length && frogs[right+1] == RIGHT_FROG){
this.move(right + 1);
}
else if(left >0 && frogs[left-1] == LEFT_FROG){
this.move(left-1);
}
else if(frogs[left] == LEFT_FROG){
this.move(left);
}
}
}
}
else if(left == -1){
if(frogs[right] == LEFT_FROG && right<length-1){
this.move(right+1);
}
else{
this.move(right);
}
}
else if(right == length){
if(frogs[left] == RIGHT_FROG && left > 0){
this.move(left-1);
}
else{
this.move(left);
}
}
}
System.out.println("step:" + step);
}

private void move(int i){
int temp = frogs[i];
frogs[i] = frogs[zeroIndex];
frogs[zeroIndex] = temp;
zeroIndex = i;
step++;
print();
}

private boolean isMoveEnd(int value){
int i=0; int max= zeroIndex;
if(value == LEFT_FROG){
i = zeroIndex+1;
max = length;
}
for(int j=i; j<max; j++){
if(frogs[j]!=value){
return false;
}
}
return true;
}

private void print(){
StringBuffer stringBuffer = new StringBuffer();
for(int frog : frogs){
if(frog>-1){
stringBuffer.append(" " +frog + "    ");
}
else{
stringBuffer.append(frog + "    ");
}

}
System.out.println(stringBuffer.toString());
}
}