Delete-occurrences-of-an-element-if-it-occurs-more-than-n-times

Description

Solutions

Julia

function deleteNth(arr::Vector, n::Int64)

    # 第一种解法：filter()
    # d = Dict{Any,Int64}()

    # function select(key)
    #     value = get(d, key, 0)
    #     d[key] = value + 1
    #     return value < n
    # end

    # return filter(select, arr)

    # 第二种解法：循环，相当于 foreach()
    d = Dict{Any,Int64}()
    output = []

    for key ∈ arr
        value = get(d, key, 0)
        if value < n
            push!(output, key)
            d[key] = value + 1
        end
    end

    return output
end

#> deleteNth (generic function with 1 method)

using Test
@test deleteNth([20, 37, 20, 21], 1) == [20, 37, 21]

#> Test Passed
#>   Expression: deleteNth([20, 37, 20, 21], 1) == [20, 37, 21]
#>    Evaluated: Any[20, 37, 21] == [20, 37, 21]

@test deleteNth([1, 1, 3, 3, 7, 2, 2, 2, 2], 3) == [1, 1, 3, 3, 7, 2, 2, 2]

#> Test Passed
#>   Expression: deleteNth([1, 1, 3, 3, 7, 2, 2, 2, 2], 3) == [1, 1, 3, 3, 7, 2, 2, 2]
#>    Evaluated: Any[1, 1, 3, 3, 7, 2, 2, 2] == [1, 1, 3, 3, 7, 2, 2, 2]

JavaScript

/**
 * 去掉已出现次数大于n的字符
 * @param {Array} arr 
 * @param {Integer} n 
 * @returns 
 */
const deleteNth = (arr, n) => {
    let map = new Map();

    return arr.filter(key => {
        let value = map.get(key) || 0; // ||左边的表达式转换成Bool为false，返回||右边的表达式
        map.set(key, value + 1);
        return map.get(key) <= n;
    });
};


console.log(deleteNth([1, 1, 3, 3, 7, 2, 2, 2, 2], 2)); // [1, 1, 3, 3, 7, 2, 2]

#> [
#>   1, 1, 3, 3,
#>   7, 2, 2
#> ]