LeetCode 47. Permutations II

最新推荐文章于 2019-09-19 16:04:00 发布

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本文介绍了一种算法，用于生成包含重复元素的集合的所有唯一排列。通过递归方式实现，避免了生成重复的全排列结果。示例代码展示了如何为输入数组 [1,1,2] 计算所有不重复的排列。

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].

#include <iostream>
#include <vector>
using namespace std;

void permutationUnique(vector<int>& nums, vector< vector<int> >& res, vector<int>& path, vector<bool>& used) {
   if(path.size() == nums.size()) {
        res.push_back(path);
        return;
    }
    for(int i = 0; i < nums.size(); ++i) {
        if(i > 0 && nums[i] == nums[i - 1] && used[i - 1]) {continue;}  <strong>// once duplciated and used, keep on going.</strong>
        if(!used[i]) {
            used[i] = true;
            path.push_back(nums[i]);
            permutationUnique(nums, res, path, used);
            used[i] = false;
            path.pop_back();
        }
    }
}

vector< vector<int> > permutationUnique(vector<int>& nums) {
    if(nums.size() == 0) return {};
    vector< vector<int> > res;
    vector<int> path;
    int start = 0;
    vector<bool> used(nums.size(), false);
    permutationUnique(nums, res, path, used);
    return res;
}

int main(void) {
    vector<int> nums{1, 1, 2};
    vector< vector<int> > res = permutationUnique(nums);
    for(int i = 0; i < res.size(); ++i) {
        for(int j = 0; j < res[0].size(); ++j) {
            cout << res[i][j] << endl;
        }
        cout << endl;
    }
}