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";

如何打印此数组中从Monkey开始到绵羊结束的所有元素？

Answer 1

您可以使用模运算符，以便您的索引变量在达到ZODIAC数组的长度时会回绕为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]); } 

另一种方法是，以slice阵列的两个部分首先进入正确的顺序：

 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个答案，但没有一个答案对我有用：

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

循环12次，并使用模运算符，以便我们可以计算4、5，... 10、11、0、1、2、3。

Answer 3

很容易，只需从数组的开始到结尾进行打印，然后从数组的开始一直打印到开始。

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

解决它的另一种有趣方式：

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

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

这会将数组的副本添加到末尾，然后仅从起始索引打印12。

Answer 5

在我看来，最直接的方法是一个do-while循环，从索引开始，然后进行包装，直到回到该索引为止。 在这种情况下，可以使用do -block使事情变得简单：

 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) 

评论：在我看来，大多数其他答案似乎都不必要地复杂（使用模），或者效率低下（复制值和新数组）。 从头到尾都可以避免，在我看来这更容易维护