# 31.next-permutation

## Statement

• Difficulty: Medium
• Tag: `数组` `双指针`

• 例如，`arr = [1,2,3]` ，以下这些都可以视作 `arr` 的排列：`[1,2,3]``[1,3,2]``[3,1,2]``[2,3,1]`

• 例如，`arr = [1,2,3]` 的下一个排列是 `[1,3,2]`
• 类似地，`arr = [2,3,1]` 的下一个排列是 `[3,1,2]`
• `arr = [3,2,1]` 的下一个排列是 `[1,2,3]` ，因为 `[3,2,1]` 不存在一个字典序更大的排列。

``````输入：nums = [1,2,3]

``````

``````输入：nums = [3,2,1]

``````

``````输入：nums = [1,1,5]

``````

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

• Difficulty: Medium
• Tag: `Array` `Two Pointers`

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 considered permutations of `arr`: `[1,2,3]`, `[1,3,2]`, `[3,1,2]`, `[2,3,1]`.

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`

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
void nextPermutation(vector<int> &nums) {
next_permutation(all(nums));
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````