如何基于起始索引在数组中循环/换行？

Question

var ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
var STARTING_ZODIAC = "MONKEY";

How can I print all the elements in this array starting with Monkey and finishing with sheep?

Answer 1

You can use the modulo operator so that your index variable wraps around to 0 once it reaches the length of the ZODIAC array:

 const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"]; const STARTING_ZODIAC = "MONKEY"; const startIndex = ZODIAC.indexOf(STARTING_ZODIAC); console.log(STARTING_ZODIAC); for (let i = startIndex + 1; i !== startIndex; i = (i + 1) % ZODIAC.length) { console.log(ZODIAC[i]); } 

Another method would be to slice the two parts of the array into the proper order first:

 const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"]; const STARTING_ZODIAC = "MONKEY"; const startIndex = ZODIAC.indexOf(STARTING_ZODIAC); [ ...ZODIAC.slice(startIndex), ...ZODIAC.slice(0, startIndex) ].forEach(str => console.log(str)); 

Answer 2

6 answers, but none of them use the obvious to me solution:

for (let i = 0; i < ZODIAC.length; i++) {
   console.log(ZODIAC[(startIndex + i) % ZODIAC.length]);
}

Loop 12 times, and use the modulus operator so we can count 4, 5, ... 10, 11, 0, 1, 2, 3.

Answer 3

Easy enough, just print from the start to the end of the array, and then from beginning of the array to the start.

function printZodiacs(startingZodiac, zodiacs) {
  const startIndex = zodiacs.indexOf(startingZodiac);

  // start to end of array
  for (let i = startIndex; i < zodiacs.length; i++) {
    console.log(zodiacs[i]);
  }

  // beginning of array to to start
  for (let i = 0; i < startIndex; i++) {
    console.log(zodiacs[i]);
  }
}

printZodiacs(STARTING_ZODIAC, ZODIAC);

Answer 4

Another fun way to solve it:

let doubleZodiac = ZODIAC.concat(ZODIAC);
let start = ZODIAC.indexOf(STARTING_ZODIAC);

for (let i = 0; i < ZODIAC.length; i++) {
  console.log(doubleZodiac[start + i]);
}

This adds a copy of the array to the end, and then just prints 12 from the starting index.

Answer 5

What seems most straightforward to me is a do-while loop, starting at the index and then wrapping until you get back to that index. This is one case where it makes sense to use a do -block to keep things simple:

 const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"]; const STARTING_ZODIAC = "MONKEY"; let i = start_ndx = ZODIAC.indexOf(STARTING_ZODIAC); do { console.log(i, ZODIAC[i++]) i == ZODIAC.length && (i=0) } while (i!==start_ndx) 

Comment: To me, most other answers seem unnecessarily complex (using modulo), or inefficient (copying values and new arrays); where as seen abvoe, both can be avoided, which in my perspective is easier to maintain