Algorithm
Problem Name: 31. Next Permutation
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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int cmp(const void *a, const void *b) {
return *(int *)a - *(int *)b;
}
void nextPermutation(int* nums, int numsSize) {
int i, j, k;
for (i = numsSize - 1; i > 0; i --) {
if (nums[i - 1] < nums[i]) {
j = i - 1; // this is the first small number from tail
while (i < numsSize - 1 &&
nums[i + 1] > nums[j]) { // find the second small number in the tail
i ++;
}
k = nums[i]; // swap it
nums[i] = nums[j];
nums[j] = k;
qsort(&nums[j + 1], numsSize - j - 1, sizeof(int), cmp);
return;
}
}
qsort(nums, numsSize, sizeof(int), cmp);
}
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Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
void nextPermutation(vector<int>& nums) {
int left = 0, right = -1;
for(int i = nums.size() - 1; i >= 0; i--)
for(int j = i - 1; j >= 0; j--)
if(nums[j] < nums[i] && (j > left || right == -1)) left = j, right = i;
if(right == -1) sort(nums.begin(), nums.end());
else{
swap(nums[left], nums[right]);
sort(nums.begin() + left + 1, nums.end());
}
}
};
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Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public void nextPermutation(int[] nums) {
int idx = nums.length - 2;
while (idx >= 0 && nums[idx + 1] < = nums[idx]) {
idx--;
}
if (idx >= 0) {
int endIdx = nums.length - 1;
while (nums[endIdx] < = nums[idx]) {
endIdx--;
}
swap(nums, idx, endIdx);
}
reverse(nums, idx + 1);
}
private void reverse(int[] nums, int startIdx) {
int endIdx = nums.length - 1;
while (startIdx < endIdx) {
swap(nums, startIdx, endIdx);
startIdx++;
endIdx--;
}
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
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Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const nextPermutation = function(nums) {
const n = nums.length
let k
for(let i = n - 2; i >= 0; i--) {
if(nums[i] < nums[i + 1]) {
k = i
break
}
}
if(k == null) {
reverse(nums, 0, n - 1>
} else {
let end
for(let i = n - 1; i >= 0; i--) {
if(nums[i] > nums[k]) {
end = i
break
}
}
swap(nums, k, end)
reverse(nums, k + 1, n - 1)
}
function reverse(arr, start, end) {
while(start < end) {
swap(arr, start, end)
start++
end--
}
}
function swap(arr, i, j> {
;[arr[i], arr[j]] = [arr[j], arr[i]];
}
};
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#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def nextPermutation(self, nums):
perm, l = False, len(nums) - 2
while 0 <= l:
r = len(nums) - 1
while l < r and nums[r] <= nums[l]:
r -= 1
if r <= l:
l -= 1
else:
nums[l], nums[l + 1:], perm = nums[r], sorted(nums[l + 1:r] + [nums[l]] + nums[r + 1:]), True
break
if not perm: nums.sort()
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Output
#6 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _031_NextPermutation
{
public void NextPermutation(int[] nums)
{
int i = nums.Length - 1;
while (i > 0 && nums[i - 1] >= nums[i])
i--;
if (i < = 0)
{
Array.Reverse(nums);
return;
}
int j = nums.Length - 1;
while (j >= 0 && nums[j] < = nums[i - 1])
j--;
int temp = nums[j];
nums[j] = nums[i - 1];
nums[i - 1] = temp;
Array.Reverse(nums, i, nums.Length - i);
}
}
}
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