算法题：将一个数组旋转k步

前端算法题：将一个数组旋转k步

例如：输入一个数组 [1,2,3,4,5,6,7]

          k = 3, 即旋转3步

          输出 [5,6,7,1,2,3,4]

两种思路：

• 思路一：把末尾的元素挨个pop，然后unshift到数组前面，时间复杂度 O(n^2)

//用pop和unshift
function rotate1(arr,k){
   const length = arr.length
   if(!k || length === 0) return arr
   const step = Math.abs(k % length)//取绝对值
   for (let i = 0;i<step;i++){
       const n = arr.pop()
       arr.unshift(n)
   }
   return arr
}
const arr = [1,2,3,4,5,6,7,0]
const arr1 = rotate1(arr,3)
console.log(arr1)

• 思路二：把数组拆分，最后 conat 拼接到一起，时间复杂度 O(1)

        //用slice和concat
        function rotate2(arr,k){
            const length = arr.length
            if(!k || length === 0) return arr
            const step = Math.abs(k % length)

            const part1 = arr.slice(-step)
            const part2 = arr.slice(0,length-step)
            const part3 = part1.concat(part2)

            return part3
        }
        const arr = [1,2,3,4,5,6,7,0]
        const arr1 = rotate2(arr,3)
        console.log(arr1)