标签：ssi process blog 交换 res void cout sed oss
Permutations
Given a collection of distinct 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], [3,2,1] ]
分析： 全排列实现思想是从start开始，分别和后面的数进行交换，递归求解
class Solution {
public:
void swap(int i, int j, vector<int>& nums){
int temp =nums[i];
nums[i] = nums[j];
nums[j]= temp;
}
void allRange(int start, vector<int>& nums, vector<vector<int>>& res)
{
if(start==nums.size()-1)
return;
for(int i =start; i<nums.size(); i++){
cout << start << " "<<i<<endl;
swap(start, i, nums);
if(start!=i)
res.push_back(nums);
allRange(start+1, nums, res);
swap(i, start, nums);
}
}
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> res;
if(nums.size()==0)
return res;
res.push_back(nums);
allRange(0, nums, res);
return res;
}
};
标签：ssi process blog 交换 res void cout sed oss
原文地址：http://www.cnblogs.com/willwu/p/6227608.html
