LeetCode JavaScript 30 Days Challenge – Day7 – 2629. Function Composition

    AllenLi
    LeetCode JavaScript 30 Days ChallengeLeetCode Js-ArrayResult

    Given an array of functions [f1, f2, f3, …, fn], return a new function fn that is the function composition of the array of functions.
    The function composition of [f(x), g(x), h(x)] is fn(x) = f(g(h(x))).
    The function composition of an empty list of functions is the identity function f(x) = x.
    You may assume each function in the array accepts one integer as input and returns one integer as output.

    給予一個陣列函式 [f1, f2, f3, ..., fn]，回傳一個新的函式 fn 是這個陣列函式的函式組合。
函式組合[f(x), g(x), h(x)]為fn(x) = f(g(h(x)))。
假如為空的函式列表的函式組合，則函式恆為input的值，如f(x) = x。

你可以假設陣列中的每個函式都接受一個整數作為輸入，並回傳一個整數當作結果。

    Example 1:

    Input: functions = [x => x + 1, x => x * x, x => 2 * x], x = 4
Output: 65

    Example 2:

    Input: functions = [x => 10 * x, x => 10 * x, x => 10 * x], x = 1
Output: 1000

    Example 3:

    Input: functions = [], x = 42
Output: 42

    solution:
    這題最重要的是了解題目的說明，題目將條件依序放到陣列中，
    同時希望我們由右到左的讀取陣列函式的內容，
    這個過程我們可以使用reduceRight()來完成指定功能，如下：

    Code 1: BigO(n)

    var compose = function(functions) {
    return function (x) {
        return functions.reduceRight((acc, cur) => cur(acc), x);
    };
};

    FlowChart:
    Example 1

    Input: functions = [x => x + 1, x => x * x, x => 2 * x], x = 4
functions.reduceRight((acc, cur) => cur(acc), x)

functions.reduceRight((acc, cur) => cur(2 * 4), 4) //acc = 8
functions.reduceRight((acc, cur) => cur(8 * 8), 8) //acc = 64
functions.reduceRight((acc, cur) => cur(64 + 1), 64) //acc = 65
return 65;

    Example 2

    Input: functions = [x => 10 * x, x => 10 * x, x => 10 * x], x = 1
functions.reduceRight((acc, cur) => cur(acc), x)

functions.reduceRight((acc, cur) => cur(10 * 1), 1) //acc = 10
functions.reduceRight((acc, cur) => cur(10 * 10), 10) //acc = 100
functions.reduceRight((acc, cur) => cur(10 + 100), 100) //acc = 1000
return 1000;

    Example 3

    Input: functions = [], x = 42
functions.reduceRight((acc, cur) => cur(acc), x)

functions.reduceRight((acc, cur) => cur(acc), 42)
return 42;

    Code 2: BigO(n)

    var compose = function(functions) {
    return function (x) {
        return functions.reverse().reduce((acc, cur) => cur(acc), x);
    };
};

    Code 3: BigO(n)

    var compose = function(functions) {
	return function(x) {
        while (functions.length) {
            x = functions.pop()(x);
        }
        return x;
    }
};

    Code 4: BigO(n)

    var compose = function(functions) {
    return function recursion(x) {      
        //如果funcitons中沒有陣列元素，則呼叫temp匿名函式回傳x的值。  
        if (functions.length === 0) return x;

        //*重點，將x重新賦值為：移除functions中最右邊的陣列的同時，帶入(x)的參數進去執行取得結果。     
        x = functions.pop()(x);      

        //重複呼叫遞迴函式直到陣列中無元素。  
        return recursion(x);    
    }
};

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