Java实现：一维矩阵位置转二维矩阵(x,y)坐标

Great question! Let's break this down with clear, generalizable rules—since the core logic depends on how the 2D matrix was flattened into a 1D array (most often row-first, which matches your 10×10 example).

Key Assumptions First #

First, let's clarify two common indexing conventions (critical for getting the right coordinates):

• 0-based indexing: Used in almost all programming languages (e.g., Python, C, Java), where the first element is at position 0.
• 1-based indexing: Used in math, spreadsheets, or some legacy systems, where the first element is at position 1.

Case 1: Row-Major (Row-First) Order (Most Common) #

This is the flattening method your 10×10 example uses: we store all elements of row 0 first, then row 1, and so on until the last row.

For 0-Based Indexing #

Given:

• ROW: Number of rows in the original 2D matrix
• COL: Number of columns in the original 2D matrix
• pos: Position in the 1D array (starts at 0)

The 2D coordinates (x, y) (where x = row index, y = column index) are:

x = pos // COL  # Integer division (discard remainder)
y = pos % COL   # Modulo operation (get remainder)

Example: 11×12 Matrix (0-Based)

If pos = 25:

• x = 25 // 12 = 2 (3rd row, since we start counting at 0)
• y = 25 % 12 = 1 (2nd column)
  Corresponding 2D coordinate: (2, 1)

For 1-Based Indexing #

If your 1D position starts at 1, adjust first to 0-based, then convert back:

pos_0 = pos - 1
x = (pos_0 // COL) + 1
y = (pos_0 % COL) + 1

Example: 11×12 Matrix (1-Based)

If pos = 25:

• pos_0 = 24
• x = (24 // 12) + 1 = 2 + 1 = 3
• y = (24 % 12) + 1 = 0 + 1 = 1
  Corresponding 2D coordinate: (3, 1)

Case 2: Column-Major (Column-First) Order #

Less common in general programming, but used in languages like Fortran or some linear algebra libraries. Here, we store all elements of column 0 first, then column 1, etc.

For 0-Based Indexing #

x = pos % ROW
y = pos // ROW

For 1-Based Indexing #

pos_0 = pos - 1
x = (pos_0 % ROW) + 1
y = (pos_0 // ROW) + 1

Verify Your 10×10 Example #

Your 10×10 solution (x = pos / 10, y = pos % 10) aligns perfectly with row-major 0-based indexing—since integer division of pos by 10 (the number of columns) gives the row index, and modulo 10 gives the column index.

内容的提问来源于stack exchange，提问作者Huy Nguyen