算法设计与应用基础：第十七周（2）

 

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90. Subsets II

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• Total Accepted: 108212
• Total Submissions: 305272
• Difficulty: Medium
• Contributor: LeetCode

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

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解题思路：采取每次取一个元素或者不取的方法（前提是先对原数组排序），处理有重复元素的方法就是判断每次要取的元素是否和之前一个元素相等，如果相等就continue，因为这个元素无论取不取都已经包含在上一个元素的情况中了（剪枝),代码如下

class Solution {  public:      vector<vector<int> > subsetsWithDup(vector<int> &S) {          // Start typing your C/C++ solution below          // DO NOT write int main() function                      vector<int> path;          vector<vector<int> > result;            sort(S.begin(), S.end());          sub(S, 0, path, result);          return result;      }        void sub(vector<int> &s, int begin, vector<int> &path, vector<vector<int> > &result) {          result.push_back(path);            for (int i = begin; i < s.size(); ++i) {              if (i != begin && s[i] == s[i - 1]) continue;                path.push_back(s[i]);              sub(s, i + 1, path, result);              path.pop_back();          }      }  }; 

补充：对于有重复元素的处理方法，排序判断是否进行循环。