40. Combination Sum II \ 46. Permutations \ 47. Permutations II

40. Combination Sum II

class Solution {public:    void combinationSumBT(vector<vector<int> > & rel, int stt, int &last, vector<int> &candidates, vector<int> &tmp, int target) {        if(target == 0) {            rel.push_back(tmp);            return;        }        if(stt == last || target < 0)            return;         for(int i = stt; i < last; i++) {            if(target >= 0) {                if(candidates[i] > target) return;                if(i&&candidates[i]==candidates[i-1]&&i > stt) continue;                tmp.push_back(candidates[i]);                combinationSumBT(rel, i+1, last, candidates, tmp, target - candidates[i]);                tmp.pop_back();            }         }    }    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        int c_len = candidates.size();        vector<vector<int> > rel;        vector<int> tmp;        sort(candidates.begin(), candidates.end());        combinationSumBT(rel, 0, c_len, candidates, tmp, target);        return rel;    }};

46. Permutations

题目描述：

Given a collection of distinct numbers, return all possible permutations.
For example,
[1,2,3] have the following permutations:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]
Subscribe to see which companies asked this question.

class Solution {public:    void permutation(vector<vector<int>> &rel, vector<int> &nums, int stt, int last, vector<int> &tmp) {        if(stt == last) {            rel.push_back(nums);            return;        }        for(int i = stt; i < last; i++) {            swap(nums[stt], nums[i]);            permutation(rel, nums, stt+1, last, tmp);            swap(nums[stt], nums[i]);        }    }    vector<vector<int>> permute(vector<int>& nums) {        int nums_len = nums.size();        vector<vector<int>> rel;        vector<int> tmp;        permutation(rel, nums, 0, nums_len, tmp);        return rel;    }};

47. Permutations II

题目描述

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],  [2,1,1]]

解决方法

法一：使用set去掉重复部分。

class Solution {public:    void permutation2(set<vector<int> > &rel, int stt, const int &last, vector<int>&nums) {        if(stt == last) {            rel.insert(nums);            return;        }        for(int i = stt; i < last; i++) {            swap(nums[i], nums[stt]);            permutation2(rel, stt+1, last, nums);            swap(nums[i], nums[stt]);        }    }    vector<vector<int>> permuteUnique(vector<int>& nums) {        int nums_len = nums.size();        set<vector<int>> tmp;        permutation2(tmp, 0, nums_len, nums);        vector<vector<int> > rel(tmp.begin(), tmp.end());        return rel;    }};

这种做法速度慢，而且浪费空间。所以正确的做法应该是在回溯的部分使用剪枝去除冗余。

法二：
纯粹调用函数：

public:    vector<vector<int>> permuteUnique(vector<int>& nums) {        auto beg = nums.begin();        auto end = nums.end();        vector<vector<int>> ret;        sort(beg, end);        do {            ret.push_back(nums);        } while (next_permutation(beg, end));        return ret;    }};

法三：击败91%的代码：

class Solution {public:    vector<vector<int> > permuteUnique(vector<int> &num) {        int n = num.size();        vector<vector<int>> res;          sort(num.begin(), num.end());    //sort the list        permuteUnique(num, res, n, 0);        return res;    }    void permuteUnique(vector<int> &num, vector<vector<int>> &res, int n, int s) {        if (s == n) {            res.push_back(num);             return;        }        for (int j = s; j < n; j++) {            if (j > s & num[j] == num[j-1]) continue;    //prevent duplicates            move(num, j, s);    //set the s-th element in the permutation to be                                 //num[j], while leaving the rest elements sorted            permuteUnique(num, res, n, s+1);            move(num, s, j);    //reset        }    }    void move(vector<int> &num, int j, int i) {         num.insert(num.begin() + i + (i > j), num[j]);        num.erase(num.begin() + j + (j > i));    }};

法四：击败92.94%。这里需要注意的是在函数permutation2里面，如果last使用&的话会比较慢，变成46%。

class Solution {public:    void permutation2(vector<vector<int> > &rel, int stt, int last, vector<int>&nums) {        if(stt == last) {            rel.push_back(nums);            return;        }        for(int i = stt; i < last; i++) {            if(i > stt & nums[i] == nums[i-1]) continue;            move(nums, i, stt);//swap(nums[stt], nums[i]);            permutation2(rel, stt+1, last, nums);            move(nums, stt, i);//swap(nums[stt], nums[i]);        }    }    void move(vector<int> &num, int j, int i) {         num.insert(num.begin() + i + (i > j), num[j]);        num.erase(num.begin() + j + (j > i));    }    vector<vector<int>> permuteUnique(vector<int>& nums) {        int nums_len = nums.size();        vector<vector<int> > rel;        sort(nums.begin(), nums.end());        permutation2(rel, 0, nums_len, nums);        return rel;    }};

法五：使用普通的DFS计算。击败92.94%的代码

class Solution {public:    void permutation2(vector<vector<int> > &rel, int stt, const int last, vector<int>nums) {        if(stt == last) {            rel.push_back(nums);            return;        }        for(int i = stt; i < last; i++) {            if(i != stt && nums[i] == nums[stt]) continue;            swap(nums[stt],  nums[i]);            permutation2(rel, stt+1, last, nums);        }    }    vector<vector<int>> permuteUnique(vector<int>& nums) {        int nums_len = nums.size();        vector<vector<int> > rel;        sort(nums.begin(), nums.end());        permutation2(rel, 0, nums_len, nums);        return rel;    }};