LeetCode 题解（106）: Permutations

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题目：

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1],[3,1,2], and [3,2,1].

题解：

典型的Backtracing or Depth First Search。要注意Java和Python的传引用与C++的不同。

C++版：

class Solution {public:    vector<vector<int>> permute(vector<int>& nums) {        vector<vector<int>> results;        if(nums.size() == 0)            return results;        vector<int> result;        vector<bool> used(nums.size(), false);        trace(results, result, used, nums);        return results;    }        void trace(vector<vector<int>>& results, vector<int>& result, vector<bool>& used, vector<int>& nums) {        if(result.size() == used.size()) {            results.push_back(result);            return;        }        for(int i = 0; i < nums.size(); i++) {            if(!used[i]) {                used[i] = true;                result.push_back(nums[i]);                trace(results, result, used, nums);                result.pop_back();                used[i] = false;            }        }    }};

Java版：

public class Solution {    public List<List<Integer>> permute(int[] nums) {        List<List<Integer>> results = new ArrayList<List<Integer>>();        if(nums.length == 0)            return results;        List<Integer> result = new ArrayList<>();        boolean[] used = new boolean[nums.length];        trace(results, result, used, nums);        return results;    }        public void trace(List<List<Integer>> results, List<Integer> result, boolean[] used, int[] nums) {        if(result.size() == nums.length) {            List<Integer> local = new ArrayList<>();            local.addAll(result);            results.add(local);            return;        }        for(int i = 0; i < nums.length; i++) {            if(!used[i]) {                used[i] = true;                result.add(nums[i]);                trace(results, result, used, nums);                result.remove(result.size()-1);                used[i] = false;            }        }    }}

Python版：

class Solution:    # @param {integer[]} nums    # @return {integer[][]}    def permute(self, nums):        if len(nums) == 0:            return []        used = [False for i in range(len(nums))]        results = []        result = []                def trace(results, result, used, nums):            if len(result) == len(nums):                temp = result[:]                results.append(temp)                return            for i in range(len(nums)):                if not used[i]:                    used[i] = True                    result.append(nums[i])                    trace(results, result, used, nums)                    result.pop()                    used[i] = False                            trace(results, result, used, nums)        return results

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