47. Permutations II

weixin_30379911

于 2018-10-17 19:46:00 发布

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CC 4.0 BY-SA版权

文章标签： c/c++ runtime

原文链接：http://www.cnblogs.com/ruruozhenhao/p/9806601.html

本文介绍了一种使用C++解决包含重复元素的集合所有唯一排列的方法。通过递归和排序，避免了重复的排列结果，实现了高效求解。代码经实际测试，运行速度优于46.26%的在线提交。

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

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

AC code:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> v;
        sort(nums.begin(), nums.end());
        recursion(nums, v, 0, nums.size()-1);
        return v;
    }

    void recursion(vector<int> nums, vector<vector<int>>& v, int i, int j) {
        if (i == j) {
            v.push_back(nums);
            return;
        }
        for (int k = i; k <= j; ++k) {
            if (i != k && nums[i] == nums[k])
                continue;
            swap(nums[i], nums[k]);
            recursion(nums, v, i+1, j);
        }
    }
};

Runtime: 28 ms, faster than 46.26% of C++ online submissions for Permutations II.

转载于:https://www.cnblogs.com/ruruozhenhao/p/9806601.html