Given an array, rotate the array to the right by k steps, where k is non-negative.
<!--more-->
Example 1:
代码语言：Swift复制Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]Example 2:
代码语言：Swift复制Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]ReverseArray
解法1：
循环弹出末尾元素，放入数组第一个位置，每弹出一次 k-1，直到 k=0截止
代码语言：Swift复制class Solution {
func rotate(_ nums: inout [Int], _ k: Int) {
// Solution 1
var times = k
while (times > 0) {
let last = nums.removeLast()
nums.insert(last, at: 0)
times -= 1
}
}能达到最终结果，但是跟算法无法，没有 reverseArrray
解法2：
reverseArray 的逻辑是：比如有nums=1, 2, 3, 4, 5, 6, 7, key = 3，则对1, 2, 3, 4 rotate得到4, 3, 2, 1，对5, 6, 7 rotate 得到7, 6, 5，然后合并得到4, 3, 2, 1, 7, 6, 5，再 rotate 得到5, 6, 7, 1, 2, 3, 4
代码如下：
代码语言：Swift复制class Solution {
func rotate(_ nums: inout [Int], _ k: Int) {
// Solution 2
let count = nums.count
let tempK = k % count
reverse(&nums, 0, count - tempK - 1)
reverse(&nums, count - tempK, count-1)
reverse(&nums, 0, count-1)
}
func reverse(_ nums: inout [Int], _ i: Int, _ j: Int) {
var mutI = i
var mutJ = j
while (mutI < mutJ) {
nums.swapAt(mutI, mutJ)
mutI = 1
mutJ -= 1
}
}
}
