46. Permutations, 47. Permutations II

Given a collection of distinct 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],
[3,2,1]
]

无重复数据的排列组合：
isVisited 的前后变化很重要

vector<vector<int>> permute(vector<int>& nums) {    vector<vector<int>> res;    vector<int> tempres;    vector<bool> isVisited(nums.size(), false);    helper(nums, res, tempres, isVisited);    return res;}void helper(vector<int>& nums, vector<vector<int>>& res, vector<int>& tempres, vector<bool>& isVisited) {    if (tempres.size() == nums.size()) {        res.push_back(tempres);        return;    }    for (int i = 0; i < nums.size(); i++) {        if (isVisited[i] == true) continue;        tempres.push_back(nums[i]);        isVisited[i] = true;        helper(nums, res, tempres, isVisited);        tempres.pop_back();        isVisited[i] = false;    }    return;}

1. permutation II
   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],
[2,1,1]
]

有重复数据的排列组合，必须要事前sort一下，然后保证同一个位置上枚举的数不是同一个。

vector<vector<int>> permuteUnique(vector<int>& nums) {    sort(nums.begin(), nums.end());    vector<vector<int>> res;    vector<int> tempres;    vector<bool> isVisited(nums.size(), false);    helper(nums, res, tempres, isVisited);    return res;}void helper(vector<int>& nums, vector<vector<int>>& res, vector<int>& tempres, vector<bool>& isVisited) {    if (tempres.size() == nums.size()) {        res.push_back(tempres);        return;    }    int lastnum = nums[0] - 1;    for (int i = 0; i < nums.size(); i++) {        if (isVisited[i] == true || nums[i] == lastnum) continue;        tempres.push_back(nums[i]);        isVisited[i] = true;        lastnum = nums[i];        helper(nums, res, tempres, isVisited);        tempres.pop_back();        isVisited[i] = false;    }    return;}