排列问题

Leetcode 46. Permutations

给定一个包含不同整数的集合，生成所有可能的排列。

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> result;
        help(nums, 0, result);
        return result;
    }
    
    // permute num[begin..end]
    // invariant: num[0..begin-1] have been fixed/permuted
    void help(vector<int>& nums, int begin, vector<vector<int>>& result){
        if(begin >= nums.size()){
            // one permutation instance
            result.push_back(nums);
            return;
        }
        for(int i = begin; i < nums.size(); ++i){
            swap(nums[i], nums[begin]);
            help(nums, begin + 1, result);
            // reset
            swap(nums[i], nums[begin]);
        }
    }
};

 Leetcode 47. Permutations II

给定一个可能包含重复整数的集合，生成所有可能的排列。

分析：值传递？

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> result;
        sort(nums.begin(), nums.end());
        help(nums, 0, result);
        return result;
    }
    
    void help(vector<int> nums, int begin, vector<vector<int>>& result){
        if(begin == nums.size() - 1){
            result.push_back(nums);
            return;
        }
        for(int i = begin; i < nums.size(); ++i){
            if(i != begin && nums[i] == nums[begin]) continue;
            else{
                swap(nums[begin], nums[i]);
                help(nums, begin + 1, result);
            }
        }
    }
};

 Leetcode 31. Next Permutation

实现next permutation

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int n = nums.size(), k, l;
        for(k = n - 2; k >= 0; --k){
            if(nums[k] < nums[k + 1])
                break;
        }
        if(k < 0) reverse(nums.begin(), nums.end());
        else{
            for(l = n - 1; l > k; --l){
                if(nums[l] > nums[k])
                    break;
            }
            swap(nums[k], nums[l]);
            reverse(nums.begin() + k + 1, nums.end());
        }
    }
};

 Leetcode 60. Permutation Sequence

集合$[1,2,3,...,n]$可以构成$n!$个排列。给定$n,k$，求第k个排列（index from 1).

分析：每一个数字开头的排列共有$(n - 1)!$个，确定第k个在哪一组，把那个数字放到s[0]上，然后同样的方法确定s[1...n-1].

class Solution {
public:
    string getPermutation(int n, int k) {
        int i, j, f = 1;
        // left part of s is partially formed permutation, right part is the leftover chars.
        string s(n, '0');
        for(int i = 1; i <= n; ++i){
            f *= i;
            s[i - 1] += i; // make s become 1234...n
        }
        for(i = 0, k--; i < n; ++i){
            f /= n - i;
            j = i + k / f; // calculate index of char to put at s[i]
            char c = s[j];
            // remove c by shifting to cover up (adjust the right part).
            for(; j > i; --j)
                s[j] = s[j - 1];
            k %= f;
            s[i] = c;
        }
        return s;
    }
};