笨鸟

天道酬勤,思者常新;博观约取,厚积薄发;心如止水,气贯长虹;淡泊明志,宁静致远。
posts - 10, comments - 0, trackbacks - 0, articles - 1
问题描述如下:
    “一个20*20的数组如下所示
         08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
         49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
         81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
         52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
         22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
         24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
         32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
         67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
         24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
         21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
         78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
         16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
         86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
         19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
         04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
         88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
         04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
         20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
         20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
         01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
    在某一方向上的四个数如红色所示,98*78*00*62=0.
    求给出20*20数组的任何方向(上,下,左斜,右斜)四个数相乘的最大值。"


    代码如下:

/**
     * 求给出20*20数组的任何方向四个数相乘的最大值
     
*/

    
private static int getGreastestNumberBy(int[][] n) {
        
int max = 0;
        
for (int i = 0; i < n.length; i++{
            
for (int j = 0; j < n[i].length; j++{
                
if (j < 3 && i < n.length - 3{
                    
if (max < n[i][j] * n[i][j + 1* n[i][j + 2* n[i][j + 3]) {
                        max 
= n[i][j] * n[i][j + 1* n[i][j + 2* n[i][j + 3];
                    }

                    
if (max < n[i][j] * n[i + 1][j] * n[i + 2][j] * n[i + 3][j]) {
                        max 
= n[i][j] * n[i + 1][j] * n[i + 2][j] * n[i + 3][j];
                    }

                    
if (max < n[i][j] * n[i + 1][j + 1* n[i + 2][j + 2]
                            
* n[i + 3][j + 3]) {
                        max 
= n[i][j] * n[i + 1][j + 1* n[i + 2][j + 2]
                                
* n[i + 3][j + 3];
                    }

                }
 else if (j > n[i].length - 4 && i < n.length - 3{
                    
if (max < n[i][j] * n[i + 1][j] * n[i + 2][j] * n[i + 3][j]) {
                        max 
= n[i][j] * n[i + 1][j] * n[i + 2][j] * n[i + 3][j];
                    }

                    
if (max < n[i][j] * n[i + 1][j - 1* n[i + 2][j - 2]
                            
* n[i + 3][j - 3]) {
                        max 
= n[i][j] * n[i + 1][j - 1* n[i + 2][j - 2]
                                
* n[i + 3][j - 3];
                    }

                }
 else if (j < n[i].length - 3 && i > n.length - 4{
                    
if (max < n[i][j] * n[i][j + 1* n[i][j + 2* n[i][j + 3]) {
                        max 
= n[i][j] * n[i][j + 1* n[i][j + 2* n[i][j + 3];
                    }

                }
 else if (j < n[i].length - 4 && i < n.length - 4{
                    
if (max < n[i][j] * n[i][j + 1* n[i][j + 2* n[i][j + 3]) {
                        max 
= n[i][j] * n[i][j + 1* n[i][j + 2* n[i][j + 3];
                    }

                    
if (max < n[i][j] * n[i + 1][j] * n[i + 2][j] * n[i + 3][j]) {
                        max 
= n[i][j] * n[i + 1][j] * n[i + 2][j] * n[i + 3][j];
                    }

                    
if (max < n[i][j] * n[i + 1][j + 1* n[i + 2][j + 2]
                            
* n[i + 3][j + 3]) {
                        max 
= n[i][j] * n[i + 1][j + 1* n[i + 2][j + 2]
                                
* n[i + 3][j + 3];
                    }

                    
if (max < n[i][j] * n[i + 1][j - 1* n[i + 2][j - 2]
                            
* n[i + 3][j - 3]) {
                        max 
= n[i][j] * n[i + 1][j - 1* n[i + 2][j - 2]
                                
* n[i + 3][j - 3];
                    }

                }
 
            }

        }

        
return max;
    }
    得到值为70600674

    请不吝赐教。
    @anthor CLumsyBirdZ


-----------------------------
博观约取,厚积薄发


只有注册用户登录后才能发表评论。


网站导航: