如何通过递归减少传播

Question

这里是for循环方法，我想要的结果与下面的代码相同

function reduce (dollars) {
  let arr = [];
  let change = [10000,5000,2000,1000,500,200,100,50,20,10,1];

  if(dollars === 0) return arr
  for(key of change){
    if(key <= dollars) return arr.concat(key).concat(reduce (dollars -= key));
  }
}

console.log(reduce(900))
console.log(reduce(9000))

如何以递归方式使用reduce和传播语法， 我遇到了一些错误，当我递归使用reduce 时，concat 函数被调用为无效 这是代码，

function reduce(dollar) {
  let change = [10000,5000,2000,1000,500,200,100,50,20,10,1];
  if(dollar === 0) return c
  let c = change.reduce((ar, el) => {
    if(dollar >= el) {
      return ar.concat(el).concat(reduce(dollar-=el))
    }
  }, [])
}
let [ one, two, three] = [10020,9000,900];
console.log(reduce(one));
console.log(reduce(two));
console.log(reduce(three));

reduce(one) 的输出是[ 10000, 20 ]

Answer 1

确保reduce处理程序总是返回数组，看到处理程序期望ar是一个数组。如果在reduce 处理程序中返回undefined，它将作为ar 传递到下一个reduce 迭代，这就是您遇到错误的原因。

另外，请确保在返回之前定义了c，或者如下所示修改代码以避免完全需要c：

function reduce(dollar) {

  let change = [10000,5000,2000,1000,500,200,100,50,20,10,1];
  
  /* if dollar === 0 then return [] */
  if (dollar === 0) return [];      
    
  /* Return reduced result (ie "c" in original code) */
  return change.reduce((ar, el) => {
        
    /*
    While el deducted from dollar yields positive value,
    */
    while (dollar - el >= 0) {
      
      /*
      then decrement dollar amout by current el and,
      */
      dollar -= el;  
      
      /*
      add el to current result
      */
      ar = ar.concat(el)
    }
    
    /*
    Return resulting ar array
    */
    return ar;
    
  }, [])
}

let [ one, two, three ] = [ 10020, 9000, 900 ];
console.log(reduce(one));
console.log(reduce(two));
console.log(reduce(three));