本文介绍了数组 - 在序列中查找缺少的数字的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试找到一种简单的方法来循环(迭代)数组以找到序列中所有缺失的数字,该数组看起来有点像下面的数字。
I'm trying to find an easy way to loop (iterate) over an array to find all the missing numbers in a sequence, the array will look a bit like the one below.
var numArray = [0189459,0189460,0189461,0189463,0189465];
For上面的数组我需要 0189462
和 0189464
退出。
For the array above I would need 0189462
and 0189464
logged out.
更新:这是我在Soufiane的回答中使用的确切解决方案。
UPDATE : this is the exact solution I used from Soufiane's answer.
var numArray = [0189459, 0189460, 0189461, 0189463, 0189465];
var mia= [];
for(var i = 1; i < numArray.length; i++)
{
if(numArray[i] - numArray[i-1] != 1)
{
var x = numArray[i] - numArray[i-1];
var j = 1;
while (j<x)
{
mia.push(numArray[i-1]+j);
j++;
}
}
}
alert(mia) // returns [0189462, 0189464]
更新
这是使用.reduce的整洁版本
Here's a neater version using .reduce
var numArray = [0189459, 0189460, 0189461, 0189463, 0189466];
var mia = numArray.reduce(function(acc, cur, ind, arr) {
var diff = cur - arr[ind-1];
if (diff > 1) {
var i = 1;
while (i < diff) {
acc.push(arr[ind-1]+i);
i++;
}
}
return acc;
}, []);
console.log(mia);
推荐答案
如果您知道这些数字已经排序并且正在增加:
If you know that the numbers are sorted and increasing:
for(var i = 1; i < numArray.length; i++) {
if(numArray[i] - numArray[i-1] != 1) {
//Not consecutive sequence, here you can break or do whatever you want
}
}
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