# 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 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`

1 2 7 4 3 1

1 3 1 2 4 7

1 2 7 4 3 1

1 2 7 4 3 1

1 3 7 4 2 1

1 3 1 2 4 7

``````class Solution {
public:
void nextPermutation(vector<int> &num) {
int i, j, n = num.size();
for (i = n - 2; i >= 0; --i) {
if (num[i + 1] > num[i]) {
for (j = n - 1; j > i; --j) {
if (num[j] > num[i]) break;
}
swap(num[i], num[j]);
reverse(num.begin() + i + 1, num.end());
return;
}
}
reverse(num.begin(), num.end());
}
};
``````

``````class Solution {
public:
void nextPermutation(vector<int>& nums) {
int n = nums.size(), i = n - 2, j = n - 1;
while (i >= 0 && nums[i] >= nums[i + 1]) --i;
if (i >= 0) {
while (nums[j] <= nums[i]) --j;
swap(nums[i], nums[j]);
}
reverse(nums.begin() + i + 1, nums.end());
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/31

Permutations II

Permutations

Permutation Sequence

Palindrome Permutation II

Palindrome Permutation

Minimum Adjacent Swaps to Reach the Kth Smallest Number

https://leetcode.com/problems/next-permutation/

https://leetcode.com/problems/next-permutation/discuss/13921/1-4-11-lines-C%2B%2B

https://leetcode.com/problems/next-permutation/discuss/13867/C%2B%2B-from-Wikipedia

LeetCode All in One 题目讲解汇总(持续更新中…)

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