leetcode46.全排列（中等）

思路1：用n!次next_permutation()
next_permutation()的复杂度是O(n)，所以应该不会越界，当到了排列最大值时返回false;

class Solution {
public:
    bool next(vector& nums);
    vector> permute(vector& nums) {

        vector> ans;
        int n = nums.size();
        if (n == 1) {
            ans.push_back(nums);
            return ans;
        }
        sort(nums.begin(), nums.end());
        do {
            ans.push_back(nums);
        }while (next(nums));
        return ans;
    }
};
bool Solution :: next(vector& nums) {
    int i = nums.size() - 2;
    while (i >= 0 && nums[i] >= nums[i + 1]) {
        i--;
    }
    if (i == -1) return false;
    int j = nums.size() - 1;
    while (nums[j] <= nums[i]) {
        --j;
    }
    swap(nums[i], nums[j]);
    reverse(nums.begin() + i + 1, nums.end());
    return true;
}

思路2：用dfs进行全排列
易错点：记不清全排列的模板，写的时候容易忘记对flag的更新

//模板
	if (!flag[i]) {   
		flag[i] = 1; 
		a[m] = i;
		dfs(m + 1);
		flag[i] = 0; 
	}

class Solution {
public:
    void dfs(int index, vector& each, vector> &ans, vector& nums, vector &flag);
    vector> permute(vector& nums) {
        
        int n = nums.size();
        vector each(n);
        vector flag(n, 0);
        vector> ans;
        dfs(0, each, ans, nums, flag);
        return ans;
    }
};
void Solution :: dfs(int index, vector& each, vector> &ans, vector& nums, vector &flag) {
    if (index == nums.size()) {
        ans.push_back(each);
        return ;
    }
    for (int i = 0; i < nums.size(); ++i) {
        if (!flag[i]) {
            flag[i] = 1;  //不要忘记对flag更新
            each[index] = nums[i];
            dfs(index + 1, each, ans, nums, flag);
            flag[i] = 0; //不要忘记对flag更新
        }
    }
}