LeetCode(Permutations II)
2014-04-09 12:28
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题目要求:
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
and
思路:跟之前的题目不同之处是这里元素有重复, 所以先将元素排序,然后在DFS的时候先判断当前元素与前一个元素是否相同,如果相同则判断前一个元素是否已经被选,如果被选则可以选当前元素,否则不能选。
代码:
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].
思路:跟之前的题目不同之处是这里元素有重复, 所以先将元素排序,然后在DFS的时候先判断当前元素与前一个元素是否相同,如果相同则判断前一个元素是否已经被选,如果被选则可以选当前元素,否则不能选。
代码:
class Solution { public: vector<vector<int> > ans; void DFS(vector<int>& num, int begin, vector<int>& res, vector<bool>& flag) { if(begin == num.size()) { ans.push_back(res); return ; } for (size_t i = 0; i < num.size(); ++i) { if(!flag[i])//当前元素没有被选 { //如果前面一个元素没有被选,且和当前元素相同则为避免重复则不选这个元素 if (i > 0 && num[i - 1] == num[i] && !flag[i - 1]) continue; flag[i] = true; res.push_back(num[i]); DFS(num, begin + 1, res, flag); flag[i] = false; res.pop_back(); } } } vector<vector<int> > permuteUnique(vector<int> &num) { ans.clear(); if(num.size() == 0) return ans; vector<bool> flag(num.size(), false); vector<int> res; sort(num.begin(), num.end());//一定要先排序 DFS(num, 0, res, flag); return ans; } };
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