Algorithem_ReverseArray

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

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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:

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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

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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：

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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
        }
    }
}