您的位置:首页 > 其它

Permutations

2015-08-01 14:35 369 查看


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]
.
思路:运用递归求解,第一位和后面的每一位替换,递归到最后一位。

class Solution {
public:
void permutation(vector<vector<int>> &ret, vector<int> &nums,int pStr,int pbegin)
{
if(pbegin == nums.size())
ret.push_back(nums);
else
{
for(int i = pbegin;i < nums.size();++i)
{
int tmp = nums[pbegin];
nums[pbegin] = nums[i];
nums[i] = tmp;
permutation(ret,nums,pStr,pbegin+1);
tmp = nums[pbegin];
nums[pbegin] = nums[i];
nums[i] = tmp;
}
}
}
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> ret;
if(0 == nums.size())
return ret;
permutation(ret,nums,0,0);
return ret;
}
};
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: