46. Permutations (Back-Track,Sort)

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].

思路：遍历数组，对于该字母，它可选择与它之后的字母交换或者是不交换=>带回溯的递归

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        result.clear();
        dfs(num,);
        return result;

    }
    void dfs(vector<int> num, int depth)
    {
        if(depth == num.size()-)
        {
            result.push_back(num);
            return;
        }
        dfs(num,depth+);
        int temp = num[depth];
        for(int i = depth+;i< num.size(); i++)
        {
            num[depth] = num[i];
            num[i] = temp;
            dfs(num,depth+);
            num[i] = num[depth];
            num[depth] = temp;
        }
    }

private:
    vector<vector<int> >  result;
};

思路II：

当字符串长度为2时 a1a2 a2a1
当字符串长度为3时 a3a1a2 a1a3a2 a1a2a3 a3a2a1 a2a3a1 a2a1a3
比较可以得到 其实就是把a3(多出来的元素)插在长度为2时的两个字符串的任意位置

时间复杂度：三个for循环 O(n3)

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        int size = nums.size();
        int resultSize;
        int resultIndex;
        vector<vector<int>> result;
        vector<int> resultItem(,nums[]);
        result.push_back(resultItem);
        for(int i = ; i <size; i++){ //nums[i] is the num to insert
            resultSize = result.size(); //resultSize in the preceeding insert iterate
            for(int j = ; j < resultSize; j++){ //iterate the array to do insertion
                result[j].push_back(nums[i]);
                resultIndex = j;
                for(int k = i-; k >=; k--){ //like insertion sort, adjust forward
                    result.push_back(result[resultIndex]);
                    result[result.size()-][k+] = result[resultIndex][k];
                    result[result.size()-][k] = result[resultIndex][k+];
                    resultIndex = result.size()-;
                }
            }
        }
        return result;
    }
};