我正在尝试编写一个函数，给定一个数组和n，返回数组重复不超过n次的元素。我无法改变数组的顺序。
以下是我到目前为止的代码。令我困惑的是，它适用于给定数组中的大多数元素，但不适用于其他元素。我试图找到代码不起作用的元素的押韵或原因。
function deleteNth(arr,n){
arr.forEach(function (item, index) {
var count = 0;
for (var i = 0; i < arr.length; i++) {
if (arr[i] === item) {
count++;
while (count > n) {
var remove = arr.lastIndexOf(item);
arr.splice(remove, 1);
count--;
}
}
}
});
return arr;
}
var x = deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35,
35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41,
35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7,
2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10], 2);
console.log(x);
目前返回此...
[7, 26, 21, 41, 43, 2, 26, 24, 10, 10, 10, 24, 35, 35, 43, 41, 7, 21,
41, 2, 43, 28]
但我应该得到这个......
[7, 26, 21, 41, 43, 2, 26, 24, 10, 10, 24, 35, 35, 43, 41, 7, 21, 2,
28]
对于我出错的地方的任何见解都将深表感谢。
答案 0 :(得分：4)
你放置while循环的逻辑是错误的，你需要把它放在for循环之外。
function deleteNth(arr, n) {
arr.forEach(function(item, index) {
var count = 0;
for (var i = 0; i < arr.length; i++) {
if (arr[i] === item) {
count++;
}
}
while (count > n) {
var remove = arr.lastIndexOf(item);
arr.splice(remove, 1);
count--;
}
});
return arr;
}
var x = deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35,
35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41,
35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7,
2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10
], 2);
console.log(x);
为什么呢？因为当你进行循环并从中移除东西时，你会把东西放回去。因此，当你有两个并排的物品时，你将第一个物品移开，第二个物品向下移动一个点以填充刚刚移除的物品。 i不会更改，因此您不会检查刚填补空白的项目。
我该怎么办？我会跟踪这些项目，如果我没有超过最大值附加它。
function cleanUp (arr, max) {
const cnts = {} // keep track of what we find
return arr.reduce((a, i) => { // loop over the array index by index
cnts[i] = (cnts[i] || 0) + 1; // mark that I seen the number
if (cnts[i] <= max) { // check to see if we are under the max
a.push(i) //if we are, add it to an arry
}
return a // return the array for reduce
}, [])
}
console.log(cleanUp([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35,
35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41,
35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7,
2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10], 2))
答案 1 :(得分：2)
对于快速变异版本，您可以使用单个while循环，用于计算项目的哈希表以及在拼接发生时调整索引。
function deleteNth(array, n) {
var counter = Object.create(null),
i = 0, v;
while (i < array.length) {
v = array[i];
if (!counter[v]) {
counter[v] = 0;
}
if (++counter[v] > n) {
array.splice(i, 1);
continue;
}
i++;
}
return array;
}
console.log(deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35, 35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41, 35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7, 2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10], 2));.as-console-wrapper { max-height: 100% !important; top: 0; }
更好的方法是使用过滤器并返回一个新数组。
function deleteNth(array, n) {
var counter = Object.create(null);
return array.filter(v => (counter[v] = (counter[v] || 0) + 1) <= n);
}
console.log(deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35, 35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41, 35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7, 2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10], 2));.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 2 :(得分：2)
此代码有效：
function deleteNth(arr,n){
var rem = new Array(), new_arr = new Array();
arr.forEach(function (item, index) {
if(!rem[item]) rem[item]=0;
if(rem[item]<n){
new_arr.push(item);
rem[item]++;
}
});
return new_arr;
}
console.log(deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35, 35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41, 35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7, 2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10], 2));
答案 3 :(得分：2)
你的所有代码都是真的。只需将while带出for循环。
function deleteNth(arr, n) {
arr.forEach(function(item, index) {
var count = 0;
for (var i = 0; i < arr.length; i++) {
if (arr[i] === item) {
count++;
}
}
while (count > n) {
var remove = arr.lastIndexOf(item);
arr.splice(remove, 1);
count--;
}
});
return arr;
}
var x = deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35,
35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41,
35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7,
2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10
], 2);
console.log(x);
答案 4 :(得分：2)
我喜欢Nina的过滤器（也可能表现得更好）但你也可以使用reduce：
function deleteNth(arr,n){
return arr.reduce(
([result,map],item)=>{
const count = (map.get(item)||0)+1;
return [
//do not add if more than n of this item have been added already
(count<=n)?result.concat(item):result,
map.set(item,count)//set the new count for this item and return map
]
},
[[],new Map()]//initial value for result and map
)[0];
}
以下是使用过滤器和Map的示例：
function deleteNth(arr,n){
const map = new Map();
return arr.filter(
item=>{
const count = (map.get(item)||0)+1;
map.set(item,count);
return (count<=n);
}
);
}
console.log(deleteNth([1,2,3,2,4,2,5], 2));
答案 5 :(得分：2)
如果你真的想这样做，那么这个答案不适合你。 （我认为有很好的理由使用不可变数据，但如果你想改变，其他答案之一就应该这样做。
这是一个解决方案，可以随时查看每个项目的计数，并过滤掉我们经常看到的项目：
const deleteNth = (arr, n) => {
const found = new Map()
return arr.filter(val => {
found.set(val, (found.get(val) || 0) + 1)
return found.get(val) <= n
})
}
const result = deleteNth([7, 26, 21, 41, 43, 2, 26, 24, 10, 26, 10, 10, 24, 35, 35, 35, 43, 26, 41, 7, 24, 24, 21, 24, 10, 35, 10, 7, 24, 7, 35, 26, 41, 35, 2, 43, 24, 2, 41, 26, 41, 7, 7, 26, 2, 10, 43, 10, 35, 41, 24, 7, 2, 2, 7, 2, 26, 24, 26, 43, 43, 21, 10, 28, 10], 2)
console.log(result)
另一个注意事项：如果您选择，它可能会提供更好的API：
deleteNth = (n) => (arr) => { /* ... */ }
这样你就可以只传递重复计数并返回一个过滤数组的新函数。
（另外，对于删除第n个之后所有重复值的内容，这听起来不是一个好名字。）