[LeetCode] Permutations II

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问题：

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], and [2,1,1].

分析：

这道题跟上一道题相似（点击这里）。但出现了重复元素这件事我们得小心一点。处理的方法是：每一次跟后面的元素swap之前，先检查一下后面这个元素之前是不是出现过。如果出现过，则continue；否则才交换。

代码（O(n!)）：

class Solution {public:bool already_swaped(int index, int i, const vector<int> &num) {for (int j = index; j < i; j ++) {if (num[j] == num[i])return true;}return false;}vector<vector<int> > permuteUnique(vector<int> &num) {vector<vector<int> > result;if (num.size() == 0)return result;permuteHelper(num, result, 0);return result;}void permuteHelper(vector<int> &num, vector<vector<int> > & result, int index) {if (index == num.size()) {result.push_back(num);return;}for (int i = index; i < num.size(); i ++) {if (already_swaped(index, i, num)) continue;int temp = num[i];num[i] = num[index];num[index] = temp;permuteHelper(num, result, index + 1);num[index] = num[i];num[i] = temp;}}};

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