1 2 3
4 5 6
7 8 9
这是我的正常数组,但我需要像对话一样对话
1 2 4
3 5 7
6 8 9
这是让它工作的非常愚蠢的方法,但即使它不起作用,因为我无法找到第二列元素。
for (i = 0; i < arr.length; ++i) {
for (n = 0; n < arr[0].length; ++n) {
if (i == 0 && n == 0){
arr[i][n] = 0;
} else if (i == 0 && n == 1) {
arr[i][n] = 2;
} else if (i == 1 && n == 0) {
arr[i][n] = 3;
} else if (n == 0) {
arr[i][n] = arr[i - 1][n] - arr[i - 2][n] + 1 + arr[i - 1][n];
} else {
arr[i][n] = arr[i][n - 1] - arr[i][n - 2] + 1 + arr[i][n - 1];
}
}
}
答案 0 :(得分:9)
好吧,如果您要枚举索引以便填充该模式,那么您将获得
0,0
1,0
0,1
2,0
1,1
0,2
2,1
1,2
2,2
因此,您需要遍历两个索引的 total 。即,添加剂总量。如您所见,0,0
总计0,1,0
和0,1
总计1,依此类推。给我们这样的东西:
0 1 2
1 2 3
2 3 4
要迭代这种对角线模式,我们可以执行以下操作:
// set up your matrix, any size and shape (MxN) is fine, but jagged arrays will break
int[][] matrix = {{0,0,0},{0,0,0},{0,0,0}};
// number is the value we will put in each position of the matrix
int number = 1;
// iterate while number is less than or equal to the total number of positions
// in the matrix. So, for a 3x3 matrix, 9. (this is why the code won't work for
// jagged arrays)
for (int i = 0; number <= matrix.length * matrix[0].length; i++) {
// start each diagonal at the top row and from the right
int row = 0;
int col = i;
do {
// make sure row and length are within the bounds of the matrix
if (row < matrix.length && col < matrix[row].length) {
matrix[row][col] = number;
number++;
}
// we decrement col while incrementing row in order to traverse down and left
row++;
col--;
} while (row >= 0);
}
请注意,虽然此实现适用于所有矩阵大小(和形状),但它不会尽可能高效。其中n
为matrix.length
(假设为矩阵),此实现是大O表示法中的最佳O(n^2)
类算法;但是,它有效地执行2*n^2
次迭代,而最佳解决方案只执行n^2
。
答案 1 :(得分:7)
你希望得到这样的东西:
1 2 4 7
3 5 8 B
6 9 C E
A D F G
在大小为NxN的网格中,对于网格中的每个点(x,y),您可以像这样确定值(仍然需要对偏移量进行一些修正,参见最终公式):
如果你在左上半部分,计算你上方和左边三角形的面积,并加上距离顶部的距离
如果您位于右下角(或中间),请计算您下方和右侧三角形的面积,添加距离底部的距离并从整个区域中减去
让我们尝试一下公式:
int N = 4; int[][] v = new[N][N];
for(int y = 0; y < N; y++) for(int x = 0; x < N; x++)
v[x][y] = ( x + y < N ) ?
( ( x + y + 1 ) * ( x + y ) / 2 + y + 1 ) :
( N * N + 1 - ( N - y ) - ( 2 * N - x - y - 1 ) * ( 2 * N - x - y - 2 ) / 2 );
我不知道这是多么复杂,但专家们肯定可以确认它是O(N ^ 2)?如果它有一些很酷的名称,如动态代码,请告诉我!
我在这里看到的优势是你不需要跳过内存,并且可以通过内存中的一个线性运行来填充所有字段。将其作为历史独立公式也可以由编译器优化或允许更好的并行化。如果你有一台N ^ 2单位的机器,他们可以在一次操作中计算整个矩阵。
答案 2 :(得分:4)
鉴于很多这些答案已经涵盖了基本的N×N阵列,而且有些非常有效,我继续制作了一个更强大的版本来处理M by N阵列,以及一个很好的格式化打印机,你自己的享受/自虐观察。
该方法的效率为O(N ^ 2)。打印机的格式为O(N ^ 2)。
您可以设置所需的任何行和列,假设为正整数值。
public static void main(String[] args) {
//create an M x N array
int rows = 20;
int columns = 11;
int[][] testData = new int[rows][columns];
//iteratively add numbers
int counter = 0;
for(int i = 0; i < rows; i++) {
for(int j = 0; j < columns; j++) {
testData[i][j] = ++counter;
}
}
//print our test array
printArray(testData);
System.out.println("");
//print our diagonal array
printArray(diagonal(testData));
}
此方法专门用于此示例,方法是使用M x N确定条目数,然后计算数字。如果你想根据数组中最长的项显示任何大小的数组,你可以很容易地调整这个代码来做到这一点。一个体面的挑战最好分配给读者。 O(N ^ 2)为此,但由于必须在数组中搜索最大值,所以取最大数字的数据本质上需要另一个O(N ^ 2)进行搜索。
static void printArray(int[][] array) {
//get number of digits
int count = array.length * array[0].length;
//get power of function
int power;
//probably the only time I'd ever end a for loop in a semicolon
//this gives us the number of digits we need
//You could also use logs I guess but I'm not a math guy
for(power = 0; count / Math.pow(10, power) > 1; power++);
for(int i = 0; i < array.length; i++){
System.out.print("{");
for(int j = 0; j < array[0].length; j++){
//Let's say Power is 0. That means we have a single-digit number, so we need
// +1 for the single digit. I throw in 2 to make it extra wide
System.out.print(String.format("%" + Integer.toString(power + 2)
+ "s", Integer.toString(array[i][j])));
}
System.out.println("}");
}
}
当我们考虑M x N时,有很多边缘情况要测试,所以我继续前进,似乎已经覆盖了所有这些。不是最好的,但看起来很有效。
static int[][] diagonal(int[][] input) {
//our array info
final int numRows = input.length;
final int numColumns = input[0].length;
int[][] result = new int[numRows][numColumns];
//this is our mobile index which we will update as we go through
//as a result of certain situations
int rowIndex = 0;
int columnIndex = 0;
//the cell we're currently filling in
int currentRow = 0;
int currentColumn = 0;
for(int i = 0; i < numRows; i++) {
for(int j = 0; j < numColumns; j++) {
result[currentRow][currentColumn] = input[i][j];
//if our current row is at the bottom of the grid, we should
//check whether we should roll to top or come along
//the right border
if(currentRow == numRows - 1) {
//if we have a wider graph, we want to reset row and
//advance the column to cascade
if(numRows < numColumns && columnIndex < numColumns - 1 ) {
//move current row down a line
currentRow = 0;
//reset columns to far right
currentColumn = ++columnIndex;
}
//if it's a square graph, we can use rowIndex;
else {
//move current row down a line
currentRow = ++rowIndex;
//reset columns to far right
currentColumn = numColumns - 1;
}
}
//check if we've reached left side, happens before the
//top right corner is reached
else if(currentColumn == 0) {
//we can advance our column index to the right
if(columnIndex < numColumns - 1) {
currentRow = rowIndex;
currentColumn = ++columnIndex;
}
//we're already far right so move down a row
else {
currentColumn = columnIndex;
currentRow = ++rowIndex;
}
}
//otherwise we go down and to the left diagonally
else {
currentRow++;
currentColumn--;
}
}
}
return result;
}
Input
{ 1 2 3}
{ 4 5 6}
{ 7 8 9}
{ 10 11 12}
Output
{ 1 2 4}
{ 3 5 7}
{ 6 8 10}
{ 9 11 12}
Input
{ 1 2 3 4 5 6}
{ 7 8 9 10 11 12}
{ 13 14 15 16 17 18}
{ 19 20 21 22 23 24}
{ 25 26 27 28 29 30}
{ 31 32 33 34 35 36}
Output
{ 1 2 4 7 11 16}
{ 3 5 8 12 17 22}
{ 6 9 13 18 23 27}
{ 10 14 19 24 28 31}
{ 15 20 25 29 32 34}
{ 21 26 30 33 35 36}
Input
{ 1 2 3 4 5 6}
{ 7 8 9 10 11 12}
{ 13 14 15 16 17 18}
{ 19 20 21 22 23 24}
{ 25 26 27 28 29 30}
{ 31 32 33 34 35 36}
{ 37 38 39 40 41 42}
{ 43 44 45 46 47 48}
{ 49 50 51 52 53 54}
{ 55 56 57 58 59 60}
{ 61 62 63 64 65 66}
{ 67 68 69 70 71 72}
{ 73 74 75 76 77 78}
{ 79 80 81 82 83 84}
{ 85 86 87 88 89 90}
{ 91 92 93 94 95 96}
{ 97 98 99 100 101 102}
{ 103 104 105 106 107 108}
{ 109 110 111 112 113 114}
{ 115 116 117 118 119 120}
{ 121 122 123 124 125 126}
{ 127 128 129 130 131 132}
{ 133 134 135 136 137 138}
{ 139 140 141 142 143 144}
{ 145 146 147 148 149 150}
Output
{ 1 2 4 7 11 16}
{ 3 5 8 12 17 22}
{ 6 9 13 18 23 28}
{ 10 14 19 24 29 34}
{ 15 20 25 30 35 40}
{ 21 26 31 36 41 46}
{ 27 32 37 42 47 52}
{ 33 38 43 48 53 58}
{ 39 44 49 54 59 64}
{ 45 50 55 60 65 70}
{ 51 56 61 66 71 76}
{ 57 62 67 72 77 82}
{ 63 68 73 78 83 88}
{ 69 74 79 84 89 94}
{ 75 80 85 90 95 100}
{ 81 86 91 96 101 106}
{ 87 92 97 102 107 112}
{ 93 98 103 108 113 118}
{ 99 104 109 114 119 124}
{ 105 110 115 120 125 130}
{ 111 116 121 126 131 136}
{ 117 122 127 132 137 141}
{ 123 128 133 138 142 145}
{ 129 134 139 143 146 148}
{ 135 140 144 147 149 150}
答案 3 :(得分:2)
您需要从索引0..n转换为x / y(从0到x * y)并从索引转换回x / y ...
public void toPos(int index){
return...
}
public int toIndex(int x, int y){
return...
}
我已将实施细节留给您。
答案 4 :(得分:2)
int len = 1;
int i = 1;
while(len <= arr.length){
//Fill this diagonal of length len
for(int r = 0; r < len; r++){
int c = (len - 1) - r;
arr[r][c] = i;
i++;
}
len++;
}
len--; len--;
while(len > 0){
//Fill this diagonal of length len
for(int c = arr.length - 1; c > (arr.length - len - 1); c--){
int r = arr.length - len + 2 - c;
arr[r][c] = i;
i++;
}
len--;
}
System.out.println(Arrays.deepToString(arr));
答案 5 :(得分:2)
以下是从here
转换为Java并根据您的问题进行调整的代码。
int[][] convertToDiagonal(int[][] input) {
int[][] output = new int[input.length][input.length];
int i = 0, j = 0; // i counts rows, j counts columns
int n = input.length;
for (int slice = 0; slice < 2 * n - 1; slice++) {
int z = slice < n ? 0 : slice - n + 1;
for (int k = z; k <= slice - z; ++k) {
// store slice value in current row
output[i][j++] = input[k][slice - k];
}
// if we reached end of row, reset column counter, update row counter
if(j == n) {
j = 0;
i++;
}
}
return output;
}
输入:
| 1 2 3 |
| 4 5 6 |
| 7 8 9 |
输出:
| 1 2 4 |
| 3 5 7 |
| 6 8 9 |
答案 6 :(得分:2)
这是一个简单的动态编程(ish)解决方案。你基本上可以从你最后的行动中学习。
注意: 这是O(N^2)
ALGOIRTHM
<强> 初始化: 强>
int m = 4;
int n = 4;
int[][] array = new int[m][n];;
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
array[i][j] = 0;
}
}
工作:
array[0][0] = 1;
for(int i = 0; i < m; i++){
if(i != 0){ array[i][0] = array[i-1][1]+1;}
// This is for the start of each row get 1+ the diagonal
for(int j = 1; j < n; j++){
if(i == 0){
array[i][j] = array[i][j-1]+j;
// only for the first row, take the last element and add + row Count
}else{
if(i == m-1 && j == n -1){
// This is only a check for the last element
array[i][j] = array[i][j-1]+1;
break;
}
// all middle elements: basically look for the diagonal up right.
// if the diagonal up right is out of bounds then take +2 the
// prev element in that row
array[i][j] = ((j+1) != (m)) ? array[i-1][j+1] +1: array[i][j-1]+2;
}
}
}
打印解决方案:
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
System.out.print(array[i][j]);
}
System.out.println("");
}
return 0;
}
答案 7 :(得分:1)
以下是您的问题的完整工作代码。如果您愿意,可以复制并粘贴
public class FillArray{
public static void main (String[] args){
int[][] array = {
{1,2,3},
{4,5,6},
{7,8,9}}; //This is your original array
int temp = 0; //declare a temp variable that will hold a swapped value
for (int i = 0; i < array[0].length; i++){
for (int j = 0; j < array[i].length; j++){
if (i < array.length - 1 && j == array[i].length - 1){ //Make sure swapping only
temp = array[i][j]; //occurs within the boundary
array[i][j] = array[i+1][0]; //of the array. In this case
array[i+1][0] = temp; //we will only swap if we are
} //at the last element in each
} //row (j==array[i].length-1)
} //3 elements, but index starts
//at 0, so last index is 2
}
}