31. Next Permutation

Question

A permutation of an array of integers is an arrangement of its members into a sequence or linear order.

For example, for arr = [1,2,3] , the following are all the permutations of arr : [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1] .
The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).
For example, the next permutation of arr = [1,2,3] is [1,3,2] .
Similarly, the next permutation of arr = [2,3,1] is [3,1,2] .
While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.
Given an array of integers nums , find the next permutation of nums .
The replacement must be in place and use only constant extra memory.
Example 1:

Input: nums = [1,2,3] Output: [1,3,2]

Example 2:

Input: nums = [3,2,1] Output: [1,2,3]

Example 3:

Input: nums = [1,1,5] Output: [1,5,1]

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 100

Solutions

see 3 Types of Permutation Generation
hint: lexicographical comparison :
if a[i] > b[i] then a is smaller than b
if a[i] < b[i] then a is greater than b

step 1 : scan from right to left to find the right most index i where nums[i] < nums[i+1]
e.g. [1 4 3 2] , i is at 1
step 2 : find the element to swap with pivot. From the right end, find the right most index j such that nums[j] > nums[i]. Then swap nums[i] and nums[j]
step 3 : reverse the subarray from i + 1 to the end

void nextPermutation(vector<int>& nums) {
	int n = static_cast<int>(nums.size());
	int i = n - 2;
	// step 1
	while (i >= 0 && nums[i] >= nums[i+1]) {
		i--;
	}
	
	if (i < 0) {
		// already the last permutation, just reverse it.
		reverse(nums.begin(), nums.end());
		return;
	}
	
	// step 2
	int j = n - 1;
	while (nums[i] >= nums[j]) {
		j--;
	}
	// step 3 : swap
	swap(nums[i], nums[j]);
		
	// step 4: reverse [i + 1, end)
	reverse(nums.begin() + i + 1, nums.end());
}