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].
Solution:
由于数据中有重复，在递归时需要判断是否可以交换。基本的枚举思路是每个数分别出现在第k位，然后排列剩下的元素。这里第k位的元素在枚举过程中需要判断是否已经出现过。
class Solution { public: vector<vector<int> > ans; vector<int> src; int srcLen; void perm(int i) { if(i == srcLen - 1) { vector<int> tmp; for(int i = 0;i < srcLen;i++) tmp.push_back(src[i]); ans.push_back(tmp); } else { set<int> dic; for(int k = i;k < srcLen;k++) { if(dic.find(src[k]) == dic.end()) { int t = src[i]; src[i] = src[k]; src[k] = t; perm(i + 1); src[k] = src[i]; src[i] = t; dic.insert(src[k]); } } } } vector<vector<int> > permuteUnique(vector<int> &num) { src = num; srcLen = num.size(); perm(0); return ans; } };
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原文：http://www.cnblogs.com/changchengxiao/p/3613188.html