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# 31.Next-Permutation

## 题目描述

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place and use only constant extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

## 代码

### Approach #1

• Find first decreasing element `i`
• Find number `nums[j]` greater than `nums[i]`
• swap
• reverse `i + 1` 之后的所有元素
class Solution {
public void nextPermutation(int[] nums) {
// 1. find the first decreasing element
int i = nums.length - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
i--;
}
// 2. find an element greather than nums[i]
if (i >= 0) {
int j = nums.length - 1;
while (j >= 0 && nums[j] <= nums[i]) {
j--;
}
// 3. swap nums[i] and nums[j]
swp(nums, i, j);
}
// 4. reverse element from i + 1
reverse(nums, i + 1)
}
private void reverse(int[] nums, int start) {
int i = start, j = nums.length - 1;
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}