# 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<int> previousPermuation(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 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<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: A list of integers * @return: A list of integers that's previous permuation */ public ArrayList<Integer> previousPermuation(ArrayList<Integer> 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<Integer> 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)