46、47-Permutations

難度：medium

題目描述

46-Permutations I:

47-Permutations II:

算法分析

46:
直接使用函數next_permutation(_vector.begin(), _vector.end())
47:
在46題的基礎上，使用二層循環，對得到的全排列兩兩進行比較，刪除重複的排列

代碼實現

46：

//  use function next_permutation
class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> result;
        sort(nums.begin(), nums.begin()+nums.size());
        do {
            result.push_back(nums);
        } while (next_permutation(nums.begin(), nums.begin() + nums.size()));
        return result;
    }
};

47：

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> result;
        sort(nums.begin(), nums.begin()+nums.size());
        do {
            result.push_back(nums);
        } while (next_permutation(nums.begin(), nums.begin() + nums.size()));
        int n = result.size();
        for (int i = 0; i < result.size(); ++i) {
            for (int j = i + 1; j < result.size(); ++j) {
                if (isSame(result[i], result[j]))
                    result.erase(result.begin() + j);
            }
        }
        return result;
    }
    bool isSame(vector<int> a, vector<int> b) {
        int n = a.size();
        for (int i = 0; i < n; ++i) {
            if (a[i] != b[i]) return false;
        }
        return true;
    }
};