Previous Permuation
最后更新于：2022-04-02 01:12:13
# Previous Permuation
### Source
- lintcode: [(51) Previous Permuation](http://www.lintcode.com/en/problem/previous-permuation/)
~~~
Given a list of integers, which denote a permutation.
Find the previous permutation in ascending order.
Example
For [1,3,2,3], the previous permutation is [1,2,3,3]
For [1,2,3,4], the previous permutation is [4,3,2,1]
Note
The list may contains duplicate integers.
~~~
### 题解
和前一题 [Next Permutation](http://algorithm.yuanbin.me/zh-cn/exhaustive_search/next_permutation.html) 非常类似，这里找上一个排列，仍然使用字典序算法，大致步骤如下：
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。
1. 从后往前遍历，找到第一个比`a[k]`小的数`a[l]`, 即`a[k] > a[l]`.
1. 交换`a[k]`与`a[l]`.
1. 反转`k + 1 ~ n`之间的元素。
为何不从前往后呢？因为只有从后往前才能保证得到的是相邻的排列，可以举个实际例子自行分析。
### Python
~~~
class Solution:
# @param num : a list of integer
# @return : a list of integer
def previousPermuation(self, num):
if num is None or len(num) <= 1:
return num
# step1: find nums[i] > nums[i + 1], Loop backwards
i = 0
for i in xrange(len(num) - 2, -1, -1):
if num[i] > num[i + 1]:
break
elif i == 0:
# reverse nums if reach maximum
num = num[::-1]
return num
# step2: find nums[i] > nums[j], Loop backwards
j = 0
for j in xrange(len(num) - 1, i, -1):
if num[i] > num[j]:
break
# step3: swap betwenn nums[i] and nums[j]
num[i], num[j] = num[j], num[i]
# step4: reverse between [i + 1, n - 1]
num[i + 1:len(num)] = num[len(num) - 1:i:-1]
return num
~~~
### C++
~~~
class Solution {
public:
/**
* @param nums: An array of integers
* @return: An array of integers that's previous permuation
*/
vector previousPermuation(vector &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 minimum
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& 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: A list of integers
* @return: A list of integers that's previous permuation
*/
public ArrayList previousPermuation(ArrayList nums) {
if (nums == null || 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.get(i) > nums.get(i + 1)) {
break;
} else if (i == 0) {
// reverse nums if reach minimum
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.get(i) > nums.get(j)) {
break;
}
}
// step3: swap betwenn nums[i] and nums[j]
Collections.swap(nums, i, j);
// step4: reverse between [i + 1, n - 1]
reverse(nums, i + 1, nums.size() - 1);
return nums;
}
private void reverse(List nums, int start, int end) {
for (int i = start, j = end; i < j; i++, j--) {
Collections.swap(nums, i, j);
}
}
}
~~~
### 源码分析
和 Permutation 一小节类似，这里只需要注意在step 1中`i == 0`时需要反转之以获得最大的序列。对于有重复元素，只要在 step1和 step2中判断元素大小时不取等号即可。
### 复杂度分析
最坏情况下，遍历两次原数组，反转一次数组，时间复杂度为 O(n)O(n)O(n), 使用了 temp 临时变量，空间复杂度可认为是 O(1)O(1)O(1).
### Reference
- [Permutations](http://algorithm.yuanbin.me/zh-cn/exhaustive_search/permutations.html)
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