数组 从指定长度位旋转数组

旋转数组

输入: nums = [1,2,3,4,5,6,7], k = 3 
输出: [5,6,7,1,2,3,4]

从指定的长度位，旋转数组

Swift

var newArray: [Int] = []
let array: [Int] = [10, 20, 30, 40, 50, 60]
let k: Int = 3
for i in 0..<array.count {
    let j: Int = (i + k) % array.count
    print(j, array[j])
    newArray.append(array[j])
}
print(array)
print(newArray)

输出如下：

3 40
4 50
5 60
0 10
1 20
2 30
[10, 20, 30, 40, 50, 60]
[40, 50, 60, 10, 20, 30]

newArray为旋转后的数组

将array所有元素替换

var newArray: [Int] = []
var array: [Int] = [10, 20, 30, 40, 50, 60]
let k: Int = 3
for i in 0..<array.count {
    let j: Int = (i + k) % array.count
    print(j, array[j])
    newArray.append(array[j])
}
array.replaceSubrange(0..<array.count, with: newArray)
print(array)

优化一下上面试的实现，采用一个中间数组temp

var array: [Int] = [10, 20, 30, 40, 50, 60]
let temp: [Int] = array
let k: Int = 3
for i in 0..<array.count {
    let j: Int = (i + k) % array.count
    let ele = temp[j]
    print(j, ele)
    array.replaceSubrange(i..<i+1, with: [ele])
}
print(array)

函数实现如下

func rotate(_ nums: inout [Int], _ k: Int) {
    if k > 0 && k < nums.count {
        let temp = nums
        for i in 0..<nums.count {
            nums.replaceSubrange(i..<i+1, with: [temp[(i+k) % nums.count]])
        }
    }
}

调用

var nums: [Int] = [1,2,3,4,5,6,7,8]
let k: Int = 3
rotate(&nums, k)
print(nums)

输出

[4, 5, 6, 7, 8, 1, 2, 3]