JS 算法与数据结构——排序算法(上)

请写一个 min 函数，要求 min(numbers) 能返回数组 numbers 中的最小数字。

let min = (numbers)=>{
  console.log(numbers)
  if(numbers.length>2){
    return min(
    [numbers[0],min(numbers.slice(1))]
    )
  }else{
    return Math.min.apply(null,numbers)
  }
}
 
let num = min([1,2,3,4])
console.log(num)//递归思路理解运行min([1,2,3,4])
1 min([1,min([2,3,4])) #1层
2 min([1,min([2,min([3,4])]))#2层
3 min([1,min([2,3--reutrn Math.min.apply(null,[3,4])])触发else
    return 3
4 min([1,2--reutrn Math.min.apply(null,[2,3]))触发else #2层
    return 2
5 reutrn Math.min.apply(null,[1,2])触发else  #1层
    return 1

请写出一个 sort 函数，要求 sort(numbers) 能返回一个把 numbers 从小到大排列的数组（你可以添加多余的帮助函数）

let num = [4,2,1,3]
let minNum = (numbers)=>{
  if(numbers.length>2){
    return minNum(
    [numbers[0],minNum(numbers.slice(1))]
    )
  }else{
    return Math.min.apply(null,numbers)
  }
}
 
let minIndex = (numbers)=>{
  return numbers.indexOf(minNum(numbers)); 
}
 
let sort = (numbers) => {
  if(numbers.length > 2){
  let index = minIndex(numbers)
  let min = numbers[index]
  numbers.splice(index, 1)
  console.log('min--'+min,'numbers--'+numbers)
//   console.log('minCon--',[min].concat(sort(numbers)))
  return [min].concat(sort(numbers))
}else{
  return numbers[0]<numbers[1] ? numbers :
  numbers.reverse()
} }
<br>//递归思路理解
sort([4,2,1,3]) #1层
1 sort([1]+sort([4,2,3]))#2层
2 sort([1]+sort([2]+sort([4,3])))#3层
3 sort([1]+sort([2]+sort([4,3])→[3,4]))#3层 
进入else, return numbers.reverse())
4 sort([1]+sort([2]+[3,4]))#2层
5 sort([1]+[2,3,4]))#1层
5 return [1,2,3,4]