计算x到y的最小步数

public class Main { public static int solution(int x_position, int y_position) { int step=0; if(y_position < x_position){ //互换值 int temp = x_position; x_position = y_position; y_position= temp; } else if(x_position == y_position){ return 0; } int mid = (y_position-x_position)/2; int an = 1,Sn = 1; int i=0,n=1; int remain; //记录剩余距离

    while(Sn <= mid){
        an +=1;
        n++;
        Sn += an;
    }
//	printf("n=%d Sn=%d an=%d",n,Sn,an);
    Sn -= an;
    an -= 1;	//还原
    step = 2*(n-1);		//目前走的步数
    int k;
    //*2说明还可以再走一个an+1 */
    if((mid-Sn)*2 > an){
        k = an + 1;
        step += 1;		//如果中间还能走一步，还有多 
        remain = (y_position-x_position)-Sn*2 - k;
    }else if((mid-Sn)*2 == an){
        k = an;
        step += 1;		//如果中间刚好走一步 
        remain = (y_position-x_position)-Sn*2 - k;
    }
    else{
        k = an;
        remain = (y_position-x_position)-Sn*2;
    }
    
    if((y_position-x_position)%2 == 1){		//如果是奇数 
        remain += 1;
    }
//	printf(" k=%d remain=%d",k,remain) ;
    if(remain > k){
        step += 2;
    }else if(remain > 1){
        step += 1;
    }else if(remain == 0){
        //不用额外加步数 
    }
    return step;
}

public static void main(String[] args) {
    // You can add more test cases here

    System.out.println(solution(12, 6) == 4);
    System.out.println(solution(34, 45) == 6);
    System.out.println(solution(50, 30) == 8);
}

}