# Next Permutation

## Question

### Problem Statement

Given a list of integers, which denote a permutation.

Find the next permutation in ascending order.

#### Example

For [1,3,2,3], the next permutation is [1,3,3,2]

For [4,3,2,1], the next permutation is [1,2,3,4]

#### Note

The list may contains duplicate integers.

## 题解

1. 从后往前寻找索引满足 a[k] < a[k + 1], 如果此条件不满足，则说明已遍历到最后一个。
2. 从后往前遍历，找到第一个比a[k]大的数a[l], 即a[k] < a[l].
3. 交换a[k]a[l].
4. 反转k + 1 ~ n之间的元素。

### Python

class Solution(object):
def nextPermutation(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
if nums is None or len(nums) <= 1:
return
# step1: find nums[i] < nums[i + 1], Loop backwards
i = 0
for i in xrange(len(nums) - 2, -1, -1):
if nums[i] < nums[i + 1]:
break
elif i == 0:
# reverse nums if reach maximum
nums = nums.reverse()
return
# step2: find nums[i] < nums[j], Loop backwards
j = 0
for j in xrange(len(nums) - 1, i, -1):
if nums[i] < nums[j]:
break
# step3: swap betwenn nums[i] and nums[j]
nums[i], nums[j] = nums[j], nums[i]
# step4: reverse between [i + 1, n - 1]
nums[i + 1:len(nums)] = nums[len(nums) - 1:i:-1]


### C++

class Solution {
public:
/**
* @param nums: An array of integers
* @return: An array of integers that's next permuation
*/
vector<int> nextPermutation(vector<int> &nums) {
if (nums.empty() || nums.size() <= 1) {
return nums;
}
// step1: find nums[i] < nums[i + 1]
int i = 0;
for (i = nums.size() - 2; i >= 0; --i) {
if (nums[i] < nums[i + 1]) {
break;
} else if (0 == i) {
// reverse nums if reach maximum
reverse(nums, 0, nums.size() - 1);
return nums;
}
}
// step2: find nums[i] < nums[j]
int j = 0;
for (j = nums.size() - 1; j > i; --j) {
if (nums[i] < nums[j]) break;
}
// step3: swap betwenn nums[i] and nums[j]
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
// step4: reverse between [i + 1, n - 1]
reverse(nums, i + 1, nums.size() - 1);

return nums;

}

private:
void reverse(vector<int>& nums, int start, int end) {
for (int i = start, j = end; i < j; ++i, --j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
};


### Java

public class Solution {
/**
* @param nums: an array of integers
* @return: return nothing (void), do not return anything, modify nums in-place instead
*/
public void nextPermutation(int[] nums) {
if (nums == null || nums.length == 0) return;

// step1: search the first nums[k] < nums[k+1] backward
int k = -1;
for (int i = nums.length - 2; i >= 0; i--) {
if (nums[i] < nums[i + 1]) {
k = i;
break;
}
}
// if current rank is the largest, reverse it to smallest, return
if (k == -1) {
reverse(nums, 0, nums.length - 1);
return;
}

// step2: search the first nums[k] < nums[l] backward
int l = nums.length - 1;
while (l > k && nums[l] <= nums[k]) l--;

// step3: swap nums[k] with nums[l]
int temp = nums[k];
nums[k] = nums[l];
nums[l] = temp;

// step4: reverse between k+1 and nums.length-1;
reverse(nums, k + 1, nums.length - 1);
}

private void reverse(int[] nums, int lb, int ub) {
for (int i = lb, j = ub; i < j; i++, j--) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
}