Array Reduce Transformation 遍例計算

Given an integer array nums, a reducer function fn, and an initial value init, return a reduced array.

A reduced array is created by applying the following operation: val = fn(init, nums[0]), val = fn(val, nums[1]), val = fn(val, nums[2]), ... until every element in the array has been processed. The final value of val is returned.

If the length of the array is 0, it should return init.

Please solve it without using the built-in Array.reduce method.

Example 1:

Input: 
nums = [1,2,3,4]
fn = function sum(accum, curr) { return accum + curr; }
init = 0
Output: 10
Explanation:
initially, the value is init=0.
(0) + nums[0] = 1
(1) + nums[1] = 3
(3) + nums[2] = 6
(6) + nums[3] = 10
The final answer is 10.

Example 2:

Input: 
nums = [1,2,3,4]
fn = function sum(accum, curr) { return accum + curr * curr; }
init = 100
Output: 130
Explanation:
initially, the value is init=100.
(100) + nums[0]^2 = 101
(101) + nums[1]^2 = 105
(105) + nums[2]^2 = 114
(114) + nums[3]^2 = 130
The final answer is 130.

Example 3:

Input: 
nums = []
fn = function sum(accum, curr) { return 0; }
init = 25
Output: 25
Explanation: For empty arrays, the answer is always init.

Constraints:

0 <= nums.length <= 1000
0 <= nums[i] <= 1000
0 <= init <= 1000

解法:

這是在講類似Sum的實作，需遍例每個元素用他提供的fn來做操作

type Fn = (accum: number, curr: number) => number;
// pre 為當前計算值，cur 為當前元素, init = 初值
// 每次reduce 需要回傳一個計算的結果回去
const reduce = (nums: number[], fn: Fn, init: number) => nums.reduce((pre,cur)=> fn(pre, cur), init);