LeetCode-in-All
31. Next Permutation
Medium
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 ofarr:[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 <= 1000 <= nums[i] <= 100
Solution
impl Solution {
pub fn next_permutation(nums: &mut Vec<i32>) {
if nums.len() <= 1 {
return;
}
// Step 1: Find the first decreasing element from the end
let mut i = nums.len() as i32 - 2;
while i >= 0 && nums[i as usize] >= nums[(i + 1) as usize] {
i -= 1;
}
// Step 2: If the entire sequence isn't decreasing, swap the element
if i >= 0 {
let mut j = nums.len() as i32 - 1;
while nums[j as usize] <= nums[i as usize] {
j -= 1;
}
Solution::swap(nums, i as usize, j as usize);
}
// Step 3: Reverse the elements after the found element
Solution::reverse(nums, (i + 1) as usize, nums.len() - 1);
}
// Swap elements at index i and j
fn swap(nums: &mut Vec<i32>, i: usize, j: usize) {
let temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
// Reverse the elements in the slice from index i to j
fn reverse(nums: &mut Vec<i32>, mut i: usize, mut j: usize) {
while i < j {
Solution::swap(nums, i, j);
i += 1;
j -= 1;
}
}
}