Javascript中的排序方法在for循环中不起作用吗?
我正在做一个代码挑战,我试图将排序数组中的每个数字与另一个数字进行比较,因为它们不匹配。我很难完成这个挑战,因为即使在for循环中使用排序后的数组之前,控制台日志正确地打印了该数组,但一旦在for循环中使用该数组,该数组似乎会保留其原始(未排序)顺序。代码如下:Javascript中的排序方法在for循环中不起作用吗?,javascript,arrays,sorting,numbers,Javascript,Arrays,Sorting,Numbers,我正在做一个代码挑战,我试图将排序数组中的每个数字与另一个数字进行比较,因为它们不匹配。我很难完成这个挑战,因为即使在for循环中使用排序后的数组之前,控制台日志正确地打印了该数组,但一旦在for循环中使用该数组,该数组似乎会保留其原始(未排序)顺序。代码如下: function minMinMax(array) { let minAbsent = 0 // sort the array from smallest to largest let smallestToLa
function minMinMax(array) {
let minAbsent = 0
// sort the array from smallest to largest
let smallestToLargest = array.sort((a, b) => a - b)
console.log(smallestToLargest)
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = array.sort((a, b) => b - a)
const largest = largestToSmallest[0]
// use second smallest number as starting point for loop, then after each number check
// to see if that number is in the array, if not then add the value to our array
for (i = 1, j = smallestToLargest[1]; i < smallestToLargest.length; i++, j++) {
if (j !== smallestToLargest[i]) {
minAbsent = j
}
console.log(smallestToLargest[i])
console.log('this is the min absent ' + minAbsent)
}
return [smallest, minAbsent, largest]
}
函数最小值(数组){
设minAbsent=0
//将数组从最小到最大排序
让smallestolargest=array.sort((a,b)=>a-b)
console.log(最小到最大)
常数最小=最小到最大[0]
//将数组从最大到最小排序
让largestToSmallest=array.sort((a,b)=>b-a)
常数最大=最大到最小[0]
//使用第二个最小的数字作为循环的起点,然后在每次数字检查之后
//要查看该数字是否在数组中,如果不在,则将该值添加到数组中
对于(i=1,j=smallestToLargest[1];i
当第二个console.log打印时,它读回给定的数组(使用原始的数字顺序,而不是排序的顺序)。那么是什么原因呢
let descend = Object.assign([], array);
上面的行不会通过引用复制数组
function minMinMax(array) {
let minAbsent = 0;
let descend = Object.assign([], array);
// sort the array from smallest to largest
let smallestToLargest = array.sort((a, b) => a - b)
console.log(smallestToLargest)
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = descend.sort((a, b) => b - a)
const largest = largestToSmallest[0]
// use second smallest number as starting point for loop, then after each number check
// to see if that number is in the array, if not then add the value to our array
for (i = 1, j = smallestToLargest[1]; i < smallestToLargest.length; i++, j++) {
if (j !== smallestToLargest[i]) {
minAbsent = j
}
console.log(smallestToLargest[i])
console.log('this is the min absent ' + minAbsent)
}
return [smallest, minAbsent, largest]
}
函数最小值(数组){
设minAbsent=0;
让下降=Object.assign([],数组);
//将数组从最小到最大排序
让smallestolargest=array.sort((a,b)=>a-b)
console.log(最小到最大)
常数最小=最小到最大[0]
//将数组从最大到最小排序
让最大到最小=下降。排序((a,b)=>b-a)
常数最大=最大到最小[0]
//使用第二个最小的数字作为循环的起点,然后在每次数字检查之后
//要查看该数字是否在数组中,如果不在,则将该值添加到数组中
对于(i=1,j=smallestToLargest[1];i
我想这样行得通
上面的行不会通过引用复制数组
function minMinMax(array) {
let minAbsent = 0;
let descend = Object.assign([], array);
// sort the array from smallest to largest
let smallestToLargest = array.sort((a, b) => a - b)
console.log(smallestToLargest)
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = descend.sort((a, b) => b - a)
const largest = largestToSmallest[0]
// use second smallest number as starting point for loop, then after each number check
// to see if that number is in the array, if not then add the value to our array
for (i = 1, j = smallestToLargest[1]; i < smallestToLargest.length; i++, j++) {
if (j !== smallestToLargest[i]) {
minAbsent = j
}
console.log(smallestToLargest[i])
console.log('this is the min absent ' + minAbsent)
}
return [smallest, minAbsent, largest]
}
函数最小值(数组){
设minAbsent=0;
让下降=Object.assign([],数组);
//将数组从最小到最大排序
让smallestolargest=array.sort((a,b)=>a-b)
console.log(最小到最大)
常数最小=最小到最大[0]
//将数组从最大到最小排序
让最大到最小=下降。排序((a,b)=>b-a)
常数最大=最大到最小[0]
//使用第二个最小的数字作为循环的起点,然后在每次数字检查之后
//要查看该数字是否在数组中,如果不在,则将该值添加到数组中
对于(i=1,j=smallestToLargest[1];i
我认为这很好。问题在于,尽管有三个单独的指针,“数组”、“最小到最大”和“最大到最小”,但它们都引用了相同的实际数组对象。“array.sort(…)”对调用方法的数组进行排序,并返回对同一数组的引用 如果在创建largestToSmallest后尝试将smallestToLargest记录到控制台,您将看到smallestToLargest将按降序打印——因为在调用第二个array.sort时,它将按降序排序
let smallestToLargest = array.sort((a, b) => a - b)
console.log(smallestToLargest);
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = array.sort((a, b) => b - a)
console.log(smallestToLargest);
在随机阵列上进行测试,得出:
[1, 3, 14, 29, 311, 323]
[323, 311, 29, 14, 3, 1]
要保留“最小到最大”数组,您可以使用“切片(0)”克隆它
试试这个:
function minMinMax(array) {
let minAbsent = 0
// sort the array from smallest to largest
let smallestToLargest = array.sort((a, b) => a - b).slice(0);
console.log(smallestToLargest);
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = array.sort((a, b) => b - a).slice(0);
console.log(smallestToLargest);
const largest = largestToSmallest[0]
// use second smallest number as starting point for loop, then after each number check
// to see if that number is in the array, if not then add the value to our array
for (i = 1, j = smallestToLargest[1]; i < smallestToLargest.length; i++, j = smallestToLargest[i]) {
if (j !== smallestToLargest[i]) {
minAbsent = j
}
console.log(smallestToLargest[i])
console.log('this is the min absent ' + minAbsent)
}
console.log(smallestToLargest);
return [smallest, minAbsent, largest]
};
函数最小值(数组){
设minAbsent=0
//将数组从最小到最大排序
设smallestolargest=array.sort((a,b)=>a-b.slice(0);
控制台日志(从小到大);
常数最小=最小到最大[0]
//将数组从最大到最小排序
让largestToSmallest=array.sort((a,b)=>b-a.slice(0);
控制台日志(从小到大);
常数最大=最大到最小[0]
//使用第二个最小的数字作为循环的起点,然后在每次数字检查之后
//要查看该数字是否在数组中,如果不在,则将该值添加到数组中
对于(i=1,j=smallestToLargest[1];i
问题似乎在于,尽管有三个单独的指针,“数组”、“最小到最大”和“最大到最小”,但它们都引用了相同的实际数组对象。“array.sort(…)”对调用方法的数组进行排序,并返回对同一数组的引用
如果在创建largestToSmallest后尝试将smallestToLargest记录到控制台,您将看到smallestToLargest将按降序打印——因为在调用第二个array.sort时,它将按降序排序
let smallestToLargest = array.sort((a, b) => a - b)
console.log(smallestToLargest);
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = array.sort((a, b) => b - a)
console.log(smallestToLargest);
在随机阵列上进行测试,得出:
[1, 3, 14, 29, 311, 323]
[323, 311, 29, 14, 3, 1]
要保留“最小到最大”数组,您可以使用“切片(0)”克隆它
试试这个:
function minMinMax(array) {
let minAbsent = 0
// sort the array from smallest to largest
let smallestToLargest = array.sort((a, b) => a - b).slice(0);
console.log(smallestToLargest);
const smallest = smallestToLargest[0]
// sort the array from largest to smallest
let largestToSmallest = array.sort((a, b) => b - a).slice(0);
console.log(smallestToLargest);
const largest = largestToSmallest[0]
// use second smallest number as starting point for loop, then after each number check
// to see if that number is in the array, if not then add the value to our array
for (i = 1, j = smallestToLargest[1]; i < smallestToLargest.length; i++, j = smallestToLargest[i]) {
if (j !== smallestToLargest[i]) {
minAbsent = j
}
console.log(smallestToLargest[i])
console.log('this is the min absent ' + minAbsent)
}
console.log(smallestToLargest);
return [smallest, minAbsent, largest]
};
函数最小值(数组){
设minAbsent=0
//将数组从最小到最小排序