[LintCode] 带重复元素的排列

递归实现：

class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    vector<vector<int> > permuteUnique(vector<int> &nums) {
        // write your code here
        sort(nums.begin(), nums.end());
        vector<vector<int> > permutations;
        if (nums.empty()) return permutations;
        permutate(nums, 0, permutations);
        return permutations;
    }
private:
    void permutate(vector<int> nums, int start, vector<vector<int> >& permutations) {
        if (start == nums.size()) {
            permutations.push_back(nums);
            return;
        }
        for (int i = start; i < (int)nums.size(); i++) {
            if (i == start || nums[i] != nums[start]) {
                swap(nums[start], nums[i]);
                permutate(nums, start + 1, permutations);
            }
        }
    }
};

非递归实现（基于nextPermutation）：

class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    vector<vector<int> > permuteUnique(vector<int> &nums) {
        // write your code here
        vector<vector<int> > permutations;
        if (nums.empty()) return permutations;
        vector<int> copy(nums.begin(), nums.end());
        nextPermutation(nums);
        permutations.push_back(nums);
        while (nums != copy) {
            nextPermutation(nums);
            permutations.push_back(nums);
        }
        return permutations;
    }
private:
    void nextPermutation(vector<int>& nums) {
        int k = -1, n = nums.size();
        for (int i = n - 2; i >= 0; i--) {
            if (nums[i] < nums[i + 1]) {
                k = i;
                break;
            }
        }
        if (k == -1) {
            reverse(nums.begin(), nums.end());
            return;
        }
        int l;
        for (int i = n - 1; i > k; i--) {
            if (nums[i] > nums[k]) {
                l = i;
                break;
            }
        }
        swap(nums[l], nums[k]);
        reverse(nums.begin() + k + 1, nums.end());
    }
};