题目：
LeetCode46 I
Given a collection of distinct numbers, return all possible permutations. （Medium）
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]
]
LeetCode47 II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.（Medium）
For example,[1,1,2] have the following unique permutations:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
分析： 首先先用next_permutation解这两个题。用好STL。
Permutations I：
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> result;
int len = nums.size();
sort(nums.begin(), nums.end());
do {
result.push_back(nums);
} while (next_permutation(nums.begin(),nums.end()));
return result;
}
};
Permutations II：
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> result;
int len = nums.size();
sort(nums.begin(), nums.end());
do {
result.push_back(nums);
} while (next_permutation(nums.begin(),nums.end()));
return result;
}
};
当然，用STL写好的函数肯定不是面试的目的，permutation的递归方法也是经典的回溯算法题。
代码：
class Solution {
private:
vector<vector<int>> result;
void helper(vector<int>& nums, int start, int end) {
if (start == end) {
result.push_back(nums);
}
for (int i = start; i <= end; ++i) {
swap(nums[start], nums[i]);
helper(nums, start + , end);
swap(nums[start], nums[i]);
}
}
public:
vector<vector<int>> permute(vector<int>& nums) {
helper(nums, , nums.size() - );
return result;
}
};