求实现指定数值映射的Java优雅函数（无需if判断）

Nice catch on that tricky edge case with i=6! Your initial modulo approach was close, but it missed the swap between 6 and 7. Let's fix this without any messy if blocks—here are two clean, efficient ways to implement the mapping:

1. Pure Mathematical Expression (No Conditionals) #

This uses integer division to handle the two mapping patterns in one line:

int result = 6 - i + 7 * (i / 6);

How it works: #

• For i=1 to 5: Integer division i/6 equals 0, so the formula simplifies to 6 - i—perfectly matching your first five mappings (1→5, 2→4, etc.).
• For i=6 and 7: i/6 equals 1 (since integer division truncates towards zero), so we add 7 to the base formula. This gives us:
  • 6 - 6 + 7 = 7 (correct for i=6)
  • 6 - 7 + 7 = 6 (correct for i=7)

2. Concise Ternary Operator (High Readability) #

If you prefer something more intuitive at a glance, this ternary version makes the two mapping rules explicit:

int result = i <= 5 ? 6 - i : 13 - i;

How it works: #

• For values 1-5, we use 6 - i (since each pairs with a number that adds up to 6).
• For 6 and 7, we use 13 - i (since 6+7=13, swapping the two values).

Full Test Code Implementation #

Drop either of these into your test loop, and it'll pass seamlessly:

StringBuilder stb = new StringBuilder();
String expected = "1=5; 2=4; 3=3; 4=2; 5=1; 6=7; 7=6; ";
for (int i=1; i<8; i++) {
    // Pick either line below
    int result = 6 - i + 7 * (i / 6);
    // int result = i <=5 ? 6 -i : 13 -i;
    stb.append(i+"="+result+"; ");
}
assertEquals(expected, stb.toString());

The issue with your original 6 - i%7 was that it only accounted for the first five mappings—this approach didn't account for the 6↔7 swap, which follows a different pattern. Both solutions above neatly merge both patterns without any clunky conditional blocks.

内容的提问来源于stack exchange，提问作者Mister Vanderbilt