Leetcode: Permutations II

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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].

在I的基础上需要去掉重复出现的。

class Solution {public:    vector<vector<int> > permuteUnique(vector<int> &num) {        vector<vector<int> > result;        permuteUtil(result, num, 0, num.size() - 1);             return result;    }        void permuteUtil(vector<vector<int> > &result, vector<int> &num, int start, int end) {        if (start == end) {            result.push_back(num);        }        else {            for (int i = start; i <= end; ++i) {                if (!isDuplicate(num, start, i)) {                    swap(num, start, i);                    permuteUtil(result, num, start + 1, end);                    swap(num, start, i);                }            }        }    }        bool isDuplicate(vector<int> &num, int start, int target) {        for (int i = start; i < target; ++i) {            if (num[i] == num[target]) {                return true;            }        }                return false;    }        void swap(vector<int> &num, int i, int j) {        int tmp = num[i];        num[i] = num[j];        num[j] = tmp;    }};

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对应的另外一种解法，关键是如果两个数重复，只有前一个数使用了才能使用后一个数。

class Solution {public:    vector<vector<int> > permuteUnique(vector<int> &num) {        vector<vector<int> > result;                sort(num.begin(), num.end());        vector<bool> used(num.size(), false);        vector<int> perm;        permuteUtil(result, perm, num, used);             return result;        }        void permuteUtil(vector<vector<int> > &result, vector<int> & perm, const vector<int> &num, vector<bool> &used) {        if (perm.size() == num.size()) {            result.push_back(perm);            return;        }                for (int i = 0; i < num.size(); ++i) {            if (used[i] || i > 0 && num[i] == num[i-1] && !used[i-1]) {                continue;            }                        used[i] = true;            perm.push_back(num[i]);            permuteUtil(result, perm, num, used);            perm.pop_back();            used[i] = false;        }    }};

在

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