Permutations

Description:

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]

.

思路分析：

求一组数据a1,a2,..., an的全排列可按如下步骤求解：

1.求出a2,a3,...an的全排列{P1,P2, ... ,Pj};

2. 将a1插入进去，得到新的全排列{a1P1, a1P2,..., a1Pj};

3. 交换a1和ai（1<i<=n），重复执行上述1,2两步，再交换a1和ai，即可得到所有的排列组合。

Code:

//定义一个接口函数
vector<vector<int>> permute(vector<int>& nums) {
return permute(nums, 0, nums.size()-1);
}

vector<vector<int>> permute(vector<int>& nums,int lIndex, int rIndex)
{
if (lIndex <= rIndex)
{
vector< vector<int> >result, tempResult;
vector<int>temp;

if (lIndex == rIndex)
{
temp.push_back(nums[lIndex]);
result.push_back(temp);
return result;
}
else
{
for (int i = lIndex; i <= rIndex; ++i)
{
if (i != lIndex)
{//注意当i和lIndex相等时，一个数将和它本身异或，则结果一定为0
nums[i]^=nums[lIndex];
nums[lIndex]^=nums[i];
nums[i]^=nums[lIndex];
}
tempResult = permute(nums, lIndex+1, rIndex);

for (int j = 0; j < tempResult.size(); ++j)
{
tempResult[j].insert(tempResult[j].begin(),nums[lIndex]);
result.push_back(tempResult[j]);
}
//将两个元素交换回来
if (i != lIndex)
{
nums[i]^=nums[lIndex];
nums[lIndex]^=nums[i];
nums[i]^=nums[lIndex];
}
}
return result;
}
}
}