I did't know what next permutation really is but I have found some explanations on Wikipedia, and it helped.
Quick Explanation
The explanation below is grab from next permutation section on Wikipedia.
cf. https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
 Find the largest index
k
such thata[k]
<a[k+1]
. If no such index exist, the permutation is the last permutation.  Find the largest index
l
greater thank
such thata[k]
<a[l]
.  Swap the value of
a[k]
with that ofa[l]
.  Reverse the sequence from
a[k+1]
up to and including th final elementa[n]
.
Other Resources

https://www.youtube.com/watch?v=quAS1iydq7U
 Good resource to understand how it works.
Code
class Solution {
func nextPermutation(_ nums: inout [Int]) {
// I have created methods for each step
let k = findLargestK(&nums) // ①
if k >= 0 {
let l = findLargestL(&nums, k) // ②
swap(&nums, k, l) // ③
}
reverse(&nums, k + 1) // ④
}
/// Find the largest index k such that a[k] < a[k+1].
/// If no such index exist, the permutation is the last permutation.
private func findLargestK(_ nums: inout [Int]) > Int {
// We start from the back because we need to find a largest index.
// 2 for we need an extra space for k+1
var k = nums.count  2
while k >= 0 && nums[k] >= nums[k+1] {
k = 1
}
return k
}
/// Find the largest index l greater than k such that a[k] < a[l].
private func findLargestL(_ nums: inout [Int], _ k: Int) > Int {
// We start from the back because we need to find a largest index.
var l = nums.count  1
while l >= 0 && nums[k] >= nums[l] {
l = 1
}
return l
}
private func swap(_ nums: inout [Int], _ a: Int, _ b: Int) {
let temp = nums[a]
nums[a] = nums[b]
nums[b] = temp
}
private func reverse(_ nums: inout [Int], _ start: Int) {
var lower = start
var upper = nums.count  1
while lower < upper {
swap(&nums, lower, upper)
lower += 1
upper = 1
}
}
}
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