Problem11

问题描述:

In the 2020 grid below, four numbers along a diagonal line have been marked in red.

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


The product of these numbers is 26  63  78  14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?



这个题目和Problem8类似！
就是多了方向
我的思路是四个方向依次找到最大值，然后在最大值找到最大的~
最终的方法其实是public static int cout_row(int[] data){}
找到一行数据中的复合题目要求的最大值
这样就将其转换为Problem8了~

public static int cout_row(int[] data){
		int result = 0;
		int step = 4;
		int  max = 0;
		int first = 0;
		for(int i=0; i<=data.length-step; i++){
			if(result==0){
				result = 1;
				first = data[i];
				for(int j=0; j<step; j++){
					if(data[i+j]==0){
						i = i + j + step;
						result = 0;
						break;
					}else{
						result *= data[i+j];
					}
				}
				if(result > max){
					max = result;
				}
				i = i + step - 1;
			}else{
				if(data[i]==0){
					result = 0;
					continue;
				}else{
					result = (result*data[i])/first;
					first = data[i-step+1];
					if(result>max){
						max = result;
					}
				}
			}
		}
		
		return max;
	}
	
	public static int count_horizontal(int[][] data){
		int result = 0;
		int  max = 0;
		int step = 4;
		for(int i=0; i<data.length; i++){
			int first = data[i][0];
			result = 0;
			for(int j=0; j<data[i].length - step; j++){
				if(result==0){
					result = 1;
					for (int k = 0; k < step; k++) {
						if(data[i][j+k]==0){
							j = j + k + step ;
							result = 0;
							break;
						}else{
							result *= data[i][j+k];
							if(result>max){
								max = result;
							}
						}
					}
					j = j+step-1;
					first = data[i][j];
				}else{
					if(data[i][j]==0){
						result =0;
						continue;
					}else{
						result = result*data[i][j]/first;
						if(result>max)
							max = result;
						first = data[i][j];
					}
				}
			}
		}
		return max;
	}
	
	public static int count_vertical(int[][] data){
		int result = 0;
		int  max = 0;
		int step = 4;
		for(int i=0; i<data.length; i++){
			int first = data[0][i];
			result = 0;
			for(int j=0; j<data[i].length - step; j++){
				if(result==0){
					result = 1;
					for (int k = 0; k < step; k++) {
						if(data[j+k][i]==0){
							j = j + k + step ;
							result = 0;
							break;
						}else{
							result *= data[j+k][i];
							if(result>max){
								max = result;
							}
						}
					}
					j = j+step-1;
					first = data[j][i];
				}else{
					if(data[j][i]==0){
						result =0;
						continue;
					}else{
						result = result*data[j][i]/first;
						if(result>max)
							max = result;
						first = data[j][i];
					}
				}
			}
		}
		return max;
	}
	
	public static int count_leading_diagonal(int[][] data){
		int result = 0;
		int max = 0;
		int length = data[0].length;
		int step = 4;
		int[] row_data = new int[data.length];
		//主对角线单独处理
		for(int i=0; i<length; i++){
			row_data[i] = data[i][i];
		}
		result  = cout_row(row_data);
		if(result>max){
			max = result;
		}
		//处理主对角线方向上的
		for(int i=1; i<length; i++){
			row_data = new int[length-i];
			for(int j=0; j<length-i;j++){
				row_data[j] = data[j][i+j];
			}
			result  = cout_row(row_data);
			if(result>max){
				max = result;
			}
			for(int j=0; j<length-i;j++){
				row_data[j] = data[i+j][j];
			}
			result  = cout_row(row_data);
			if(result>max){
				max = result;
			}
		}
		
		return max ;
	}
	
	public static int count_Inclined_diagonal(int[][] data){
		int result = 0;
		int max = 0;
		int length = data[0].length;
		int step = 4;
		int[] row_data = new int[data.length];
		//斜对角线单独处理
		for(int i=0; i<length; i++){
			row_data[i] = data[length-i-1][length-i-1];
		}
		result  = cout_row(row_data);
		if(result>max){
			max = result;
		}
		//处理斜对角线方向上的
		for(int i=length-2; i>length-step; i--){
			row_data = new int[i+1];
			for(int j=0; j<=i;j++){
				row_data[j] = data[j][i-j];
			}
			result  = cout_row(row_data);
			if(result>max){
				max = result;
			}
			row_data = new int[i+1];
			for(int j=0; j<=i;j++){
				row_data[j] = data[i-j][j];
			}
			result  = cout_row(row_data);
			if(result>max){
				max = result;
			}
		}
		
		return max ;
	}