使用递归Javascript的数组的总和

Question

Looking for a way to solve this problem by recursing sum() . Right now, the code works, but I am supposed to call sum() more than once, and it should not mutate the input array.

var sum = function(array) {
    if(array.length === 0){
        return 0;
    }
    function add(array, i){
        console.log(array[i]);
        if(i === array.length-1){
            return array[i];
        }
        return array[i] + add(array, i+1);
    }
    return add(array, 0);
};
sum([1, 2, 3, 4, 5, 6]) //21

Answer 1

A one-liner that meets all your requirements:

var sum = function(array) {
    return (array.length === 0) ? 0 : array[0] + sum(array.slice(1));
}

// or in ES6

var sum = (array) => (array.length === 0) ? 0 : array[0] + sum(array.slice(1));

// Test cases
sum([1,2,3]); // 6

var s = [1,2,3];
sum(s); // 6
sum(s); // 6

Reasoning

• In a recursive call, you need to model your task as reduction to a base case. The simplest base case in this case is the empty array - at that point, your function should return zero.
• What should the reduction step be? Well you can model a sum of an array as the result of adding the first element to the sum of the remainder of the array - at some point, these successive calls will eventually result in a call to sum([]) , the answer to which you already know. That is exactly what the code above does.
• array.slice(1) creates a shallow copy of the array starting from the first element onwards , and no mutation ever occurs on the original array. For conciseness, I have used a ternary expression .

Breakdown:

sum([1,2,3])
-> 1 + sum([2,3])
-> 1 + 2 + sum([3])
-> 1 + 2 + 3 + sum([])
-> 1 + 2 + 3 + 0
-> 6

Answer 2

You're on the right track, but consider that sum could take an optional second argument (that defaults to zero) that indicates the position to start summing from...

function sum(array, n) {
    n ||= 0;
    if (n === array.length) {
        return 0;
    } else {
        return array[n] + sum(array, n + 1);
    }
}

Answer 3

arr = [1,2,3,4]
sum = arr.reduce((acc, curr)=> acc+curr)

Answer 4

You don't really need the add function inside your sum function just inline the function and initiate with, 0 as a starting point, or optionally check the i variable for undefined and initialize it to 0!

var sum = function(array, i) {
    if(array.length === 0){
        return 0;
    }
  console.log(array[i]);
  if(i === array.length-1){
    return array[i];
  }
  return array[i] + sum(array, i+1);
};
console.log(sum([1, 2, 3, 4, 5, 6],0)) //21

Answer 5

You have two solutions:

• you can use .reduce() method
• or perform a simple tail recursion

With reduction :

function sum(a, b) {
  return a + b;
}

const array = [1, 2, 3, 4, 5, 6];

//In our reduce, we will apply our sum function, 
//and pass the result as the next value
const res = array.reduce(sum);

With recursion :

function sumRec(array, acc = 0, index) {
    //We will update our accumulator, and increment
  // the value of our current index
  return index === array.length
  ? acc
  : sumRec(array, acc += array[index], ++index);
}

console.log(sumRec(array, 0, 0));

Personally, I find the first solution more elegant.

Answer 6

function sumArr(arr){
  if(arr.length>1){
    return arr.pop()+sumArr(arr);
  }else{
    return arr[0];
  }
}

Answer 7

 function sumNumbersRecursively(input){ if (input.length == 0){ return 0; } else{ return input.shift() + sumNumbersRecursively(input); } } console.log(sumNumbersRecursively([2,3,4]))

Answer 8

If you have to call sum more than once, then use the binary approach: split the array in half and recur on each piece. When you get to a length of 1, return the single value.

Does this work for you? I'm afraid I don't recall the JS syntax for array slices, so my recursion statement may be wrong in the details.

var sum = function(array) {
    if(array.length === 1){
        return array[0];
    }
    mid = array.length / 2
    return sum(array[0:mid-1]) + sum(array[mid:array.length-1])
};
sum([1, 2, 3, 4, 5, 6]) //21