Javascript 数组 - 在序列中查找缺失的数字
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原文地址: http://stackoverflow.com/questions/7317993/
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Arrays - Find missing numbers in a Sequence
提问by Mark Walters
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.
我试图找到一种简单的方法来循环（迭代）一个数组以查找序列中所有缺失的数字，该数组看起来有点像下面的那个。
var numArray = [0189459, 0189460, 0189461, 0189463, 0189465];
var numArray = [0189459, 0189460, 0189461, 0189463, 0189465];
For the array above I would need 0189462and 0189464logged out.
对于上面的数组，我需要0189462并0189464注销。
UPDATE: this is the exact solution I used from Soufiane's answer.
更新：这是我从 Soufiane 的回答中使用的确切解决方案。
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]
UPDATE
更新
Here's a neater version using .reduce
这是一个使用 .reduce 的更简洁的版本
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);回答by Soufiane Hassou
If you know that the numbers are sorted and increasing:
如果您知道数字已排序并递增：
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
}
}
回答by SammieFox
ES6-Style
ES6 风格
var arr = [0189459, 0189460, 0189461, 0189463, 0189465];
var [min,max] = [Math.min(...arr), Math.max(...arr)];
var out = Array.from(Array(max-min),(v,i)=>i+min).filter(i=>!arr.includes(i));
Result:?[189462, 189464]
结果:?[189462, 189464]
回答by kennebec
Watch your leading zeroes, they will be dropped when the array is interpreted-
注意你的前导零，当数组被解释时它们将被删除 -
var A= [0189459, 0189460, 0189461, 0189463, 0189465]
var A= [0189459, 0189460, 0189461, 0189463, 0189465]
(A returns [189459,189460,189461,189463,189465])
（A 返回 [189459,189460,189461,189463,189465]）
function absent(arr){
var mia= [], min= Math.min.apply('',arr), max= Math.max.apply('',arr);
while(min<max){
if(arr.indexOf(++min)== -1) mia.push(min);
}
return mia;
}
var A= [0189459, 0189460, 0189461, 0189463, 0189465]; alert(absent(A))
var A= [0189459, 0189460, 0189461, 0189463, 0189465]; 警报（缺席（A））
/* returned value: (Array) 189462,189464 */
/* 返回值：（数组）189462,189464 */
回答by OmChoudhary
To find a missing number in a sequence, First of all, We need to sort an array. Then we can identify what number is missing. I am providing here full code with some test scenarios. this code will identify only missing positive number, if you pass negative values even then it gives positive number.
要在序列中找到缺失的数字，首先，我们需要对数组进行排序。然后我们可以确定缺少哪个数字。我在这里提供了一些测试场景的完整代码。此代码将仅识别丢失的正数，如果您传递负值，即使它给出正数。
function findMissingNumber(inputAr) {
// Sort array
sortArray(inputAr);
// finding missing number here
var result = 0;
if (inputAr[0] > 1 || inputAr[inputAr.length - 1] < 1) {
result = 1;
} else {
for (var i = 0; i < inputAr.length; i++) {
if ((inputAr[i + 1] - inputAr[i]) > 1) {
result = inputAr[i] + 1;
}
}
}
if (!result) {
result = inputAr[inputAr.length - 1] + 1;
}
return result;
}
function sortArray(inputAr) {
var temp;
for (var i = 0; i < inputAr.length; i++) {
for (var j = i + 1; j < inputAr.length; j++) {
if (inputAr[j] < inputAr[i]) {
temp = inputAr[j];
inputAr[j] = inputAr[i];
inputAr[i] = temp;
}
}
}
}
console.log(findMissingNumber([1, 3, 6, 4, 1, 2]));
console.log(findMissingNumber([1, 2, 3]));
console.log(findMissingNumber([85]));
console.log(findMissingNumber([86, 85]));
console.log(findMissingNumber([0, 1000]));回答by Julian David
const findMissing = (arr) => {
const min = Math.min(...arr);
const max = Math.max(...arr);
// add missing numbers in the array
let newArr = Array.from(Array(max-min), (v, i) => {
return i + min
});
// compare the full array with the old missing array
let filter = newArr.filter(i => {
return !arr.includes(i)
})
return filter;
};
回答by Famous Ketoma
const findMissing = (numarr) => {
for(let i = 1; i <= numarr.length; i++) {
if(i - numarr[i-1] !== 0) {
console.log('found it', i)
break;
} else if(i === numarr.length) console.log('found it', numarr.length + 1)
}
};
console.log(findMissing([1,2,3,4,5,6,7,8,9,10,11,12,13,14]))
回答by Anu
Assuming that there are no duplicates
假设没有重复
let numberArray = [];
for (let i = 1; i <= 100; i++) {
numberArray.push(i);
}
let deletedArray = numberArray.splice(30, 1);
let sortedArray = numberArray.sort((a, b) => a - b);
let array = sortedArray;
function findMissingNumber(arr, sizeOfArray) {
total = (sizeOfArray * (sizeOfArray + 1)) / 2;
console.log(total);
for (i = 0; i < arr.length; i++) {
total -= arr[i];
}
return total;
}
console.log(findMissingNumber(array, 100));
回答by Sayeed roshan
Try as shown below
尝试如下图
// Find the missing number
let numArray = [0189459, 0189460, 0189461, 0189463, 0189468];
let numLen = numArray.length;
let actLen = Number(numArray[numLen-1])-Number(numArray[0]);
let allNumber = [];
for(let i=0; i<=actLen; i++){
allNumber.push(Number(numArray[0])+i);
}
[...allNumber].forEach(ele=>{
if(!numArray.includes(ele)){
console.log('Missing Number -> '+ele);
}
})
回答by Sayeed roshan
function missingNum(nums){
const numberArray = nums.sort((num1, num2)=>{
return num1 - num2;
});
for (let i=0; i < numberArray.length; i++){
if(i !== numberArray[i]){
return i;
}
}
}
console.log(missingNum([0,3,5,8,4,6,1,9,7]))
回答by DJ.
Please check below code.....
请检查下面的代码......
function solution(A) {
var max = Math.max.apply(Math, A);
if(A.indexOf(1)<0) return 1;
var t = (max*(max+1)/2) - A.reduce(function(a,b){return a+b});
return t>0?t:max+1;
}
