Problem 18

问题描述：
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)


解决方法：

public static long find_max_sum(int[][] numbers){
long result = 0;
long[][] sum = new long[numbers.length][numbers.length];
sum[0] = new long[3];Arrays.fill(sum[0], 0);
sum[0][0] = numbers[0][0];

for(int i=1; i<numbers.length; i++){
for(int j=0; j<numbers[i].length; j++){
if(j==0){
sum[i][j] = numbers[i][j] +sum[i-1][j];;
}else if(j==numbers[i].length-1){
sum[i][j] = numbers[i][j] +sum[i-1][j-1];
}else{
long sum1 = numbers[i][j] +sum[i-1][j];
long sum2 = numbers[i][j] +sum[i-1][j-1];
long large = sum1>sum2?sum1:sum2;
sum[i][j] = large;
}
}
}

for(int i=0; i<sum.length; i++){
for(int j=0; j<sum[i].length; j++){
System.out.printf("%10d", sum[i][j]);
}
System.out.println();
}

int row = sum.length-1;
result = sum[row][0];
for(int i=1; i<sum[row].length; i++){
if(result < sum[row][i])
result = sum[row][i];
}

return result;
}