全排列，字典顺序问题 ( permutations/leetcode）

46. Permutations

题目地址

https://leetcode.com/problems/permutations/

注意是distinct numbers，所以相对简单, 当然写出好的代码也是困难的，下面是ac代码（效率可能不高）

方法1：采用dfs遍历

class Solution {
private:
vector<vector<int>> ans;
vector<int> rs;
map<int,int> vis;

public:

void dfs(int len, vector<int> nums, int n)
{
if (len == n){
ans.push_back(rs);
return;
}

for (int i = 0; i < n; i++)
{
if (vis[i] == 0){
vis[i] = 1;
rs.push_back(nums[i]);
dfs(len + 1, nums, n); // dfs
rs.pop_back();
vis[i] = 0;
}
}
}

vector<vector<int>> permute(vector<int>& nums) {
int n = nums.size();
if (n == 0)
return ans;
dfs(0, nums, n);
return ans;
}
};

方法2： 全排列思路，就是以某个数开头，剩下的全排列，递归来做

class Solution {
private:
vector<vector<int>> ans;

public:

void recursion(int k,vector<int> nums, int n)
{
if (k >= n)
return;
if (k == n - 1){
/*  for (int i = 0; i < n; i++)
cout << nums[i] << " ";
cout << endl;*/
ans.push_back(nums);
}
for (int i = k; i < n; i++)
{
swap(nums[k], nums[i]);
recursion(k + 1, nums, n);
swap(nums[k], nums[i]);
}
}

vector<vector<int>> permute(vector<int>& nums) { // nums是distinct
int n = nums.size();
if (n == 0)
return ans;
recursion(0, nums, n);
return ans;
}
};

47. Permutations II

题目地址

https://leetcode.com/problems/permutations-ii/

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]
]

有重复数字，采用递归做法，显然有很多重复的swap操作，需要避免

可以参考：

陆草纯

http://www.cnblogs.com/ganganloveu/p/4161693.html

ac代码：

class Solution {
private:
vector<vector<int>> ans;
public:

void recursion(int k, vector<int> nums, int n)
{
if (k >= n)
return;
if (k == n - 1){
ans.push_back(nums);
}

sort(nums.begin() + k, nums.end()); // k之后的元素需要排序
// 两相同元素 ， 一个元素与两个相同的元素？
for (int i = k; i < n; i++)
{
if (i != k && nums[i] == nums[i-1])
continue;
swap(nums[k], nums[i]);
recursion(k + 1, nums, n);  // 递归
swap(nums[k], nums[i]);
}
}

vector<vector<int>> permuteUnique(vector<int>& nums) {
int n = nums.size();
if (n == 0)
return ans;

recursion(0, nums, n);
return ans;
}
};

31. Next Permutation

题目地址

https://leetcode.com/problems/next-permutation/

求下一个字典序

思路参考：

warmland

http://www.cnblogs.com/warmland/p/5219217.html

一个老外对该题较好的解释：

http://www.geeksforgeeks.org/find-next-greater-number-set-digits/

ac代码：

class Solution {
public:

void rev(vector<int> &nums, int left, int right){
for (int i = left; i <= (left + right) / 2; i++){
int tmp = nums[i];
nums[i] = nums[right - i + left];
nums[right - i + left] = tmp;
}
}

void nextPermutation(vector<int>& nums) {
int n = nums.size();
if (n <= 1)
return;

int pos = n - 2;
while (pos >= 0 && nums[pos] >= nums[pos + 1]){ // 递增，等号问题
pos--;
}
if (pos >= 0){
// 找到pos后面 比 nums[pos] 大的最小的一个数
int j = n - 1;
while (nums[j] <= nums[pos]){
j--;
}

// swap
int tmp = nums[j];
nums[j] = nums[pos];
nums[pos] = tmp;

// reverse pos之后的所有数
rev(nums, pos + 1, n - 1);
}
else{
rev(nums, 0, n - 1);
}
}
};