如何基於起始索引在數組中循環/換行？

Question

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

如何打印此數組中從Monkey開始到綿羊結束的所有元素？

Answer 1

您可以使用模運算符，以便您的索引變量在達到ZODIAC數組的長度時會回繞為0 ：

 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]); } 

另一種方法是，以slice陣列的兩個部分首先進入正確的順序：

 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個答案，但沒有一個答案對我有用：

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

循環12次，並使用模運算符，以便我們可以計算4、5，... 10、11、0、1、2、3。

Answer 3

很容易，只需從數組的開始到結尾進行打印，然后從數組的開始一直打印到開始。

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

解決它的另一種有趣方式：

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

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

這會將數組的副本添加到末尾，然后僅從起始索引打印12。

Answer 5

在我看來，最直接的方法是一個do-while循環，從索引開始，然后進行包裝，直到回到該索引為止。 在這種情況下，可以使用do -block使事情變得簡單：

 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) 

評論：在我看來，大多數其他答案似乎都不必要地復雜（使用模），或者效率低下（復制值和新數組）。 從頭到尾都可以避免，在我看來這更容易維護