从2D转换为1D阵列

Question

所以我正在努力将这个逻辑从2D应用程序转换为在一维数组中运行。对我来说最让人困惑的是数组以及它们的循环将如何运作。任何输入将不胜感激！

LcgRandom rand = new LcgRandom();
		
		int[][] triangle = new int[ROWS][];
		
		for (int i = 0; i < triangle.length; i++) {
			triangle[i] = new int[i + 1];
			//System.out.println(triangle[i]);
			for (int j = 0; j <= i; j++)
				triangle[i][j] = rand.next();
				//System.out.println(triangle[i]);
		}
		
		//Calculate row sums
		int[][] rowSums = new int[triangle.length][];
		
		for (int i = 0; i < rowSums.length; i++) {
			rowSums[i] = new int[triangle[i].length + 1];
			rowSums[i][0] = 0;
			for (int j = 0; j <= i; j++)
				rowSums[i][j + 1] = rowSums[i][j] + triangle[i][j];
		}
		
		//find the smallest triangle
		long minSum = 0;
		
		for (int i = 0; i < triangle.length; i++) {
			for (int j = 0; j < triangle[i].length; j++) {
				
				long curSum = 0;
				for (int k = i; k < triangle.length; k++) {
					curSum += rowSums[k][k - i + 1 + j] - rowSums[k][j];
					if (minSum>curSum)
						minSum=curSum;
					System.out.println(minSum);
				}
			}
		}
		return Long.toString(minSum);
	}

Answer 1

如果您打算以行主格式存储2D数组，即首先存储所有row1元素，然后线性存储row2元素，则需要调整索引计算。

假设您的2D数组有R行和C列，则每行都有C个元素。因此，您的索引计算如下：

array_1d[j + i*C] == array_2d[i][j] // This is true.

因此，您的1D数组中将包含R * C个元素。