Permutations,Permutations II,Combinations
这是使用DFS来解数组类题的典型题目,像求子集,和为sum的k个数也是一个类型
解题步骤:
1:有哪些起点,例如,数组中的每个元素都有可能作为起点,那么用个for循环就可以了。
2:是否允许重复组合
3:处理某个数,判断结果
4:dfs递归
5:还原现场
一: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]
.
代码:
class Solution {
void dfs(vector<int>& nums,int start,vector<vector<int>>& res,vector<int>& oneRes){ int n = nums.size(); if(start == n){
res.push_back(oneRes);
} for(int i= start;i<nums.size();++i){ if(i>start && nums[i]==nums[i-]){
continue;
} oneRes.push_back(nums[i]); swap(nums[i],nums[start]); dfs(nums,start+,res,oneRes); swap(nums[i],nums[start]); oneRes.pop_back();
}
}
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> res;
vector<int> oneRes; dfs(nums,,res,oneRes); return res;
}
};
二:Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,[1,1,2]
have the following unique permutations:[1,1,2]
, [1,2,1]
, and [2,1,1]
.
方法1.
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) { vector<vector<int>> res; vector<int> tmp(nums); sort(tmp.begin(),tmp.end()); res.push_back(tmp); while(next_permutation(tmp.begin(),tmp.end())){
res.push_back(tmp);
} return res;
}
};
方法2.
class Solution { void dfs(vector<int>& nums,int start,vector<vector<int>>& res,vector<int>& oneRes){ int n = nums.size(); if(start == n){
res.push_back(oneRes);
} for(int i= start;i<nums.size();++i){ if(i>start && nums[i]==nums[i-]){
continue;
} int selectNum = nums[i]; oneRes.push_back(selectNum); copy_backward(nums.begin()+start,nums.begin()+i,nums.begin()+i+);
nums[start] = selectNum; dfs(nums,start+,res,oneRes); copy(nums.begin()+start+,nums.begin()+i+,nums.begin()+start);
nums[i] = selectNum; oneRes.pop_back();
}
}
public:
vector<vector<int>> permuteUnique(vector<int>& nums) { vector<vector<int>> res;
vector<int> oneRes; sort(nums.begin(),nums.end()); dfs(nums,,res,oneRes); return res;
}
};
方法3.
class Solution {
public:
void dfs(vector<int> nums,int numsSize,int startPos,vector<vector<int>>& res,vector<int>& oneOfRes)
{
sort(nums.begin()+startPos,nums.end());
for(int i=startPos;i<numsSize;i++){
if(i>startPos && nums[i]==nums[i-]){
continue;
}
oneOfRes.push_back(nums[i]);
swap(nums[i],nums[startPos]);
if(oneOfRes.size()==numsSize){
res.push_back(oneOfRes);
}else{
dfs(nums,numsSize,startPos+,res,oneOfRes);
}
swap(nums[i],nums[startPos]);
oneOfRes.pop_back(); }
}
vector<vector<int>> permuteUnique(vector<int>& nums) {
// sort(nums.begin(),nums.end());
vector<vector<int>> res;
vector<int> oneOfRes;
int numsSize = nums.size();
dfs(nums,numsSize,,res,oneOfRes);
return res;
}
};
77. Combinations
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example,
If n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
class Solution {
void dfs(int n,int start,int k,int curk,vector<vector<int>>& res,vector<int>& oneRes){ if(curk == ){
res.push_back(oneRes);
return;
} for(int i=start;i<=n;++i){ oneRes.push_back(i); dfs(n,i+,k,curk-,res,oneRes); oneRes.pop_back(); }
}
public:
vector<vector<int>> combine(int n, int k) { vector<vector<int>> res;
vector<int> oneRes; dfs(n,,k,k,res,oneRes); return res;
}
};
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