递归练习
//max值 function getMaxVal(arr){ let maxVal=null; function loop(arr){ let [key1,...keys]=arr; if(!key1){return} maxVal=key1>maxVal?key1:maxVal; loop(keys); } loop(arr) return maxVal; } getMaxVal([9,12,1,3,6]) function getMaxVal1(arr){ function loop(arr){ let [key1,...keys]=arr; if(!keys.length){return key1} return key1>loop(keys)?key1:loop(keys); } return loop(arr) } getMaxVal1([9,12,1,3,6])
let myReverse=(arr)=>{ let [key1,...keys]=arr; if(!keys.length)return [key1] return [...myReverse(keys),key1] } myReverse(['a','b','c'])
面试题1： 楼梯问题 假如楼梯有n个台阶，每次可以走1个或2个台阶，请问走完这n个台阶有几种走法? 那如果有n个台阶呢？我们认真思考下就会发现，第1步的走法有两类：第一种是第一步走1个台阶，
第二种是第二步走2个台阶。所以n个台阶的走法就可以分为：走完1个台阶后的n-1种走法，加上走完2个台阶后的n-2种走法，用递归公式表示就是: function climbStairs(n) { return climbStairs(n - 1) + climbStairs(n - 2) }
climbStairs(n) = climbStairs(n-1) + climbStairs(n-2)
climbStairs(n-1) = climbStairs(n-2) + climbStairs(n-3)
··· ··· ···
climbStairs(5) = climbStairs(4) + climbStairs(3)
climbStairs(4) = climbStairs(3) + climbStairs(2)
climbStairs(3) = climbStairs(2) + climbStairs(1)
climbStairs(2) = 2
climbStairs(1) = 1
function climbStairs(n) { if (n == 1) return 1 if (n == 2) return 2 return climbStairs(n-1) + climbStairs(n-2) }
参考：https://juejin.im/post/5d7df6bff265da03b638dc97
https://www.jb51.net/article/144663.htm
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var oneDimensionArr2=function(arr){ return arr.reduce((pre,cur)=>{ if(!cur.children)return[] return (Array.isArray(cur.children)&&cur.children.length)?[...pre,cur,...this.oneDimensionArr2(cur.children)]:[...pre,cur] },[]) }
