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从数组拼接值并将拼接的值推入另一个数组

[英]Splice value from an array and push spliced value into another array

Hello and here is my problem.I have a task,i need to remove elements from an array and push them into another array and then return the array with removed elements. 您好,这是我的问题。我有一个任务,我需要从数组中删除元素,然后将它们推入另一个数组,然后返回带有删除元素的数组。 For example, in this case i have to remove all 0 values, and push them into 'a' array and then return 'a' array with there 0 values. 例如,在这种情况下,我必须删除所有0值,并将它们推入'a'数组,然后返回具有0个值的'a'数组。 I removed 0 values from the array by using splice method and loop, but i don't realy know how to push removed elemts into 'a' array, i've tried to use push method but i does not work for me. 我通过使用splice方法和循环从数组中删除了0个值,但我真的不知道如何将删除的elemts推送到'a'数组中,我试图使用push方法,但对我不起作用。 Hope you'll help me.Thank you everyone. 希望你能帮助我。谢谢大家。

 function moveZeros(array) { var a = []; for (var i = array.length - 1; i--;) { if (array[i] == "0") { array.splice(i, 1); } } return a; } moveZeros([1, 2, 0, 1, 0, 1, 0, 3, 0, 1]); 

Using push should work. 使用push应该可以。 The .splice method will return an array with the removed elements, and you can use the spread operator ... to pass it as a set of arguments to push : .splice方法将返回一个数组,其中包含已删除的元素,您可以使用spread运算符...将其作为push一组参数传递:

 function moveZeros(array) { var a = []; for (var i = array.length - 1; i >= 0; i--) { if (array[i] == "0") { a.push(...array.splice(i, 1)); } } return a; } const array = [0, 1, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 0]; console.log(moveZeros(array)); console.log(array) 

Finally, you should put the i-- as the final part of the loop so that it only executes when each iteration finishes (instead of as they begin). 最后,应将i--作为循环的最后一部分,以便仅在每次迭代完成时才执行(而不是在开始时)。 Then, change your condition to be i >= 0 so that you don't miss a zero at the front of the array. 然后,将条件更改为i >= 0这样您就不会在数组的前面错过零。

Array.splice() returns an array of removed elements, and you can use Array.concat() to add them to the a array. Array.splice()返回移除元件的阵列,并且可以使用Array.concat()将它们添加到a数组。

Notes: 笔记:

  1. Init i to array.length without the -1, since the condition i-- is checked before the execution of the loop's body. i初始化为不带-1的array.length ,因为条件i--在执行循环主体之前被检查。
  2. Since the the array is iterated from end to starts, concat a to the removed elements to maintain the original order. 由于数组是从头到尾进行迭代a ,因此将a到已删除的元素以保持原始顺序。

 function moveZeros(array) { var a = []; for (var i = array.length; i--;) { if (array[i] === 0) { a = array.splice(i, 1).concat(a); } } return a; } var result = moveZeros([1, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0]); // I've added a last 0 to the array to show that all items were removed console.log(result); 

You could iterate from the end and splice the array if necessary. 您可以从头开始迭代,并在必要时拼接数组。 This solution mutates the array . 此解决方案使array变异。

 function moveZeros(array) { var i = array.length, a = []; while (i--) { if (array[i] === 0) { a.push(array.splice(i, 1)[0]); } } return a; } var array = [1, 2, 0, 1, 0, 1, 0, 3, 0, 1]; console.log(moveZeros(array)); console.log(array); 
 .as-console-wrapper { max-height: 100% !important; top: 0; } 

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