2048逻辑分析

2048小游戏的基本操作是上下左右，每个操作的逻辑都不太一样！

这个时候，通过数组旋转的方式。将所有操作转换为向上的操作

git地址
https://gitee.com/hodgekou/golang.git

package mainimport (	"fmt")type arr [4][4]intfunc (t *arr) Print() {	for _, line := range t {		for _, number := range line {			fmt.Printf("%2d ", number)		}		fmt.Println()	}	fmt.Println()}func (t *arr) Left90() {	temp := new(arr)	for i, line := range t {		for j, number := range line {			temp[len(line)-j-1][i] = number		}	}	*t = *temp}func (t *arr) Right90() {	temp := new(arr)	for i, line := range t {		for j, number := range line {			// fmt.Println(i,j,j,len(line)-1-i)			temp[j][len(line)-i-1] = number		}	}	*t = *temp}func (t *arr) Up() {	for m := 0; m < 3; m++ {		for j := 0; j < 4; j++ {			for i := 0; i < 3; i++ {				if t[i][j] == 0 {					t[i][j] = t[i+1][j]					t[i+1][j] = 0				} else if t[i][j] == t[i+1][j] {					t[i][j] = t[i][j] + t[i+1][j]					t[i+1][j] = 0				}			}		}	}}func main() {	// t := arr{
  {0,2,2,4},{0},{},{0,2,0,4}}	t := arr{
  {2, 2, 2, 2}, {2, 2, 4, 2}, {}, {0, 2, 0, 4}}	t.Print()	//右	fmt.Println("右")	t.Left90()	t.Up()	t.Right90()	t.Print()	//左	fmt.Println("左")	t.Right90()	t.Up()	t.Left90()	t.Print()	//下	fmt.Println("下")	t.Right90()	t.Right90()	t.Up()	t.Left90()	t.Left90()	t.Print()	//上	fmt.Println("上")	t.Up()	t.Print()}