算法-排列组合|带重复元素的排列算法-排列组合|总结-深度优先

给出一个具有重复数字的列表，找出列表所有不同的排列。
您在真实的面试中是否遇到过这个题？ Yes 样例给出列表 [1,2,2]，不同的排列有：

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

挑战 【算法-排列组合|带重复元素的排列】使用递归和非递归分别完成该题。

class Solution { public: /* * @param :A list of integers * @return: A list of unique permutations */ vector> permuteUnique(vector &nums) { // write your code here vector> res; // vector visit(nums.size(), 0); // vector result; // sort nums sort(nums.begin(), nums.end()); // permute(nums, 0, result, visit, res); // return res; // permute(nums, 0, res); // return res; // itself res.push_back(nums); while (nextPermute(nums)) { res.push_back(nums); } return res; } void permute(vector& nums, int level, vector& result, vector &visit, vector> &res) { if (level == nums.size()) { res.push_back(result); return; } for (int i = 0; i < nums.size(); i++) { if (visit[i] > 0) { continue; } // skip duplicate // visit[i - 1] == 0 means the same value had been put at this level if (i > 0 && nums[i] == nums[i - 1] && visit[i - 1] == 0) { continue; } result.push_back(nums[i]); visit[i] = 1; permute(nums, level + 1, result, visit, res); visit[i] = 0; result.pop_back(); } } bool needswap(vector &nums, int begin, int end, int target) { for (int i = begin; i <= end; i++) { if (nums[i] == target) { return false; } } return true; } void permute(vector& nums, int start, vector> &res) { if (start == nums.size()) { res.push_back(nums); return; } for (int i = start; i < nums.size(); i++) { // at this start, same value had been posted if (i > start && !needswap(nums, start, i - 1, nums[i])) { continue; } swap(nums[i], nums[start]); permute(nums, start + 1, res); swap(nums[i], nums[start]); } } void reverse(vector &nums, int begin, int end) { while (begin < end) { swap(nums[begin], nums[end]); begin++; end--; } } bool nextPermute(vector& nums) { int n = nums.size() - 1; int m = n - 1; // first increase order while (m >= 0 && nums[m] >= nums[m + 1]) { m--; } if (m >= 0) { // fist larger than nums[m] while (nums[n] <= nums[m]) { n--; } swap(nums[m], nums[n]); } // move to out of if because we need to recover nums reverse(nums, m + 1, nums.size() - 1); // return if we have next permute if (m >= 0) { return true; } else { return false; } } };