19.2.23 [LeetCode 90] Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

题意

一个有重复元素的集合，求出所有不重复的子集

题解

 1 class Solution {
 2 public:
 3     void solve(vector<int>&nums, vector<vector<int>>&ans, vector<int>&now, int i) {
 4         int n = nums.size(), _i = i;
 5         for (; i < n; i++) {
 6             if (i != _i && nums[i] == nums[i - 1])continue;
 7             now.push_back(nums[i]);
 8             ans.push_back(now);
 9             solve(nums, ans, now, i + 1);
10             now.pop_back();
11         }
12     }
13     vector<vector<int>> subsetsWithDup(vector<int>& nums) {
14         sort(nums.begin(), nums.end());
15         vector<vector<int>>ans; vector<int>now;
16         ans.push_back(vector<int>());
17         solve(nums, ans, now, 0);
18         return ans;
19     }
20 };
View Code

循环的时候注意一下跳过重复元素即可

posted @ 2019-02-23 17:01  TobicYAL  阅读(117)  评论(0)  编辑  收藏  举报