47. Permutations II 1

题目：

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

链接： http://leetcode.com/problems/permutations-ii/

题解：

求全排列，但元素可能有重复。去重复就成为了关键。今天好好思考了一下dfs+回溯，比如1134，最外层就是求出第一个元素，比如 1， 2， 3，里面的嵌套dfs再负责第二，三，四个元素。去重复的方法是传递一个visited数组，把排序后相同的元素看成一个cluster，假如nums[i] == nums[i - 1]，但i-1没有被访问过，说明整个cluster不被访问，跳过整个cluster。

Time complexity - O(n!), Space Complexity - O(n)。

public class Solution {public List<List<Integer>> permuteUnique(int[] nums) {List<List<Integer>> res = new ArrayList<>();if(nums == null || nums.length == 0)return res;Arrays.sort(nums);ArrayList<Integer> list = new ArrayList<Integer>();boolean[] visited = new boolean[nums.length];dfs(res, list, nums, visited);return res;}private void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] nums, boolean[] visited) {if(list.size() == nums.length) {res.add(new ArrayList<Integer>(list));return;}for(int i = 0; i < nums.length; i++) {if(visited[i] || (i > 0 && nums[i] == nums[i - 1] && !visited[i - 1]))      //skip duplicatescontinue;if(!visited[i]) {visited[i] = true;list.add(nums[i]);dfs(res, list, nums, visited);list.remove(list.size() - 1);visited[i] = false;}}}
}

二刷：

Java:

DFS + Backtracking:

一刷写的一坨屎...这遍依然不清不楚。主要还是用了Permutation的代码，不同的地方是，我们使用了一个数组 - boolean[] visited。这个数组用来在dfs过程中记录已经访问过的值来避免计算重复。同时我们在dfs和backtracking的时候也要回溯这个数组。经过上述步骤，我们就可以避免在dfs的时候有重复了。比如输入数组为[1, 1, 1]，则这个最后的结果 {[1, 1, 1]}是在最外层被加入到res中去的。我们也要注意在遍历数组的时候，假如 visited[i]或者(i > 0 && nums[i] == nums[i - 1] && visited[i - 1])，要continue。

Time Complexity - O(n!)， Space Complexity - O(n)

public class Solution {public List<List<Integer>> permuteUnique(int[] nums) {List<List<Integer>> res = new ArrayList<>();if (nums == null || nums.length == 0) {return res;}Arrays.sort(nums);boolean[] visited = new boolean[nums.length];permuteUnique(res, new ArrayList<Integer>(), visited, nums);return res;}private void permuteUnique(List<List<Integer>> res, List<Integer> onePerm, boolean[] visited, int[] nums) {if (onePerm.size() == nums.length) {res.add(new ArrayList<>(onePerm));return;}for (int i = 0; i < nums.length; i++) {if (visited[i] || (i > 0 && nums[i] == nums[i - 1] && visited[i - 1])) {continue;}visited[i] = true;onePerm.add(nums[i]);permuteUnique(res, onePerm, visited, nums); onePerm.remove(onePerm.size() - 1);visited[i] = false;}}
}

Iterative： Using Next Permutation:

我们依然可以使用Permutation I里面使用了求next permutation的代码，完全搬移，都不用改的。时间和空间复杂度还需要好好计算一下。这里有点混。

Time Complexity - O(n!)， Space Complexity - O(n)

public class Solution {public List<List<Integer>> permuteUnique(int[] nums) {List<List<Integer>> res = new ArrayList<>();if (nums == null || nums.length == 0) {return res;}Arrays.sort(nums);Integer[] numsInt = new Integer[nums.length];for (int i = 0; i < nums.length; i++) {numsInt[i] = nums[i];}res.add(new ArrayList<>(Arrays.asList(numsInt)));while (hasNextPermutation(numsInt)) {res.add(new ArrayList<>(Arrays.asList(numsInt)));}return res;}private boolean hasNextPermutation(Integer[] nums) {for (int i = nums.length - 2; i >= 0; i--) {if (nums[i] < nums[i + 1]) {for (int j = nums.length - 1; j >= i; j--) {if (nums[j] > nums[i]) {swap(nums, i, j);reverse(nums, i + 1, nums.length - 1);return true;}}}}return false;}private void swap(Integer[] nums, int i, int j) {int tmp = nums[i];nums[i] = nums[j];nums[j] = tmp;}private void reverse(Integer[] nums, int i, int j) {while (i < j) {swap(nums, i++, j--);}}
}

三刷:

使用了和上一题一样的代码，也是next permutation的方法。

Java:

public class Solution {public List<List<Integer>> permuteUnique(int[] nums) {List<List<Integer>> res = new ArrayList<>();if (nums == null || nums.length == 0) return res;Arrays.sort(nums);do {List<Integer> permu = new ArrayList<>();for (int num : nums) permu.add(num);res.add(permu);} while (hasNextPermutation(nums));return res;}private boolean hasNextPermutation(int[] nums) {int len = nums.length;for (int i = len - 2; i >= 0; i--) {if (nums[i] < nums[i + 1]) {for (int j = len - 1; j > i; j--) {if (nums[j] > nums[i]) {swap(nums, i, j);reverse(nums, i + 1, len - 1);return true;}       }}}return false;}private void swap(int[] nums, int i, int j) {int tmp = nums[i];nums[i] = nums[j];nums[j] = tmp;}private void reverse(int[] nums, int i, int j) {while (i < j) swap(nums, i++, j--);}
}

Reference：

http://www.cnblogs.com/springfor/p/3898447.html

https://leetcode.com/discuss/25279/a-simple-c-solution-in-only-20-lines

https://leetcode.com/discuss/10609/a-non-recursive-c-implementation-with-o-1-space-cost

https://leetcode.com/discuss/18482/share-my-recursive-solution

https://leetcode.com/discuss/77245/line-python-solution-with-line-handle-duplication-beat-others

https://leetcode.com/discuss/62272/ac-python-clean-solution-108-ms

https://leetcode.com/discuss/55350/short-and-clean-java-solution

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