动态规划解决小米最终面试题第二小题

笨方法,机智的解法是调换下标，有需要了解的可以搜索相关内容

原题见下图

function xm2(n, bool = false) {
            let puke = []
            puke[0] = 0
            puke[1] = [1]
            puke[2] = [2, 1]
            if (bool) {
                puke[2] = [1, 2]
            }
            // puke[3] = [1, 3, 2, 4]
            for(let j=3; j<=n; j=j+1) {
                let rightTree = []
                let leftTree = []
                for(let i=0; i<Math.ceil(j/2); i++) {
                    rightTree.push(Math.floor(j/2)+1+i)
                }
                if (!bool) {
                    rightTree = rightTree.reverse()
                }
                leftTree = puke[Math.floor(j/2)].concat([])
                puke[j] = new Array(j)
                for(let m=0; m<j; m++){
                    if(m%2 == bool) {
                        puke[j][m] = rightTree.shift()
                    } else {
                        puke[j][m] = leftTree.shift()
                    }
                }
            }
            return puke[n]
        }
        function c(n, bool){
            console.log(xm2(n, bool))
        }