Leetcode: Permutations

Given a collection of 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]

,
and 

[3,2,1]

.

以前看MoreWindows的文章，有介绍这个，在此谢过。

class Solution {
public:
vector<vector<int> > permute(vector<int> &num) {
vector<vector<int> > result;
permuteUtil(result, num, 0, num.size() - 1);

return result;
}

void permuteUtil(vector<vector<int> > &result, vector<int> &num, int start, int end) {
if (start == end) {
result.push_back(num);
}
else {
for (int i = start; i <= end; ++i) {
swap(num, start, i);
permuteUtil(result, num, start + 1, end);
swap(num, start, i);
}
}
}

void swap(vector<int> &num, int i, int j) {
int tmp = num[i];
num[i] = num[j];
num[j] = tmp;
}
};

========================

另一种解法，纯粹的搜索：

class Solution {
public:
vector<vector<int> > permute(vector<int> &num) {
vector<vector<int> > result;
vector<bool> used(num.size(), false);
vector<int> perm;
permuteUtil(result, perm, num, used);

return result;
}

void permuteUtil(vector<vector<int> > &result, vector<int> & perm, const vector<int> &num, vector<bool> &used) {
if (perm.size() == num.size()) {
result.push_back(perm);
return;
}

for (int i = 0; i < num.size(); ++i) {
if (used[i]) {
continue;
}

used[i] = true;
perm.push_back(num[i]);
permuteUtil(result, perm, num, used);
perm.pop_back();
used[i] = false;
}
}
};