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## Problem

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

## Code

556. Next Greater Element III

4 步:

1. 从后往前遍历，找到第一个降序 A[x]
2. 从后往前遍历，找到第一个比 A[x]大的数字 A[y]
3. 交换 A[x]和 A[y]
4. 将 x 之后的数字排序
 3  2
1  3 
 3 
 2 

class Solution {
public void nextPermutation(int[] nums) {
int i = nums.length - 2;
while(i >= 0 && nums[i + 1] <= nums[i]){
i--;
}

if(i == -1){
Arrays.sort(nums);
return;
}

int j  = nums.length - 1;
while(j >= 0 && nums[j] <= nums[i]){
j--;
}

swap(nums, i, j);
reverse(nums, i + 1);
}

private void reverse(int[] nums, int i){
int left = i;
int right = nums.length - 1;
while(left < right){
swap(nums, left, right);
left++;
right--;
}
}

private void swap(int[] nums, int i, int j){
int num = nums[i];
nums[i] = nums[j];
nums[j] = num;
}
}