LeetCode： Combination Sum系列

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],

]

回溯

class Solution {
public:
void tocom(vector<vector<int>> &res,int beg,int end,int k,vector<int> &v)
{
if(!k)
{
res.push_back(v);
return;
}
for(int i=beg;i<=end;++i)
{
vector<int> v1{v};
v1.push_back(i);
tocom(res,i+1,end,k-1,v1);
}
}
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> res;
vector<int> v{};
tocom(res,1,n,k,v);
return res;
}
};

39. Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.

The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,

A solution set is:

[

[7],

[2, 2, 3]

]

class Solution {
public:
void search(vector<int>& candidates,vector<vector<int>> &res,int &i,int &target,int &sz,vector<int> v0)
{
for(int j=i;j<sz;++j)
{
int m=target-candidates[j];
if(target>candidates[j])
{
if(m>=candidates[i])
{
vector<int> v1{v0};
v1.push_back(candidates[j]);
search(candidates,res,j,m,sz,v1);
}
}
else if(!m)
{
vector<int> v1{v0};
v1.push_back(candidates[j]);
res.push_back(v1);//出现能够符合条件的加入结果
return;
}
else
{
return;
}
}
}
//总结：不要随意在结果集里面添加元素，想着以后再删，这样只会拉长时间，不如只添加合乎要求的。
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> res;
sort(candidates.begin(),candidates.end());//可以及时停止循环
int sz=candidates.size();
for(int i=0;i<sz;++i)
{
int m=target-candidates[i];
if(target>candidates[i])
{
if(m>=candidates[i])
{
search(candidates,res,i,m,sz,{candidates[i]});
}
}
else if(!m)
{
res.push_back({candidates[i]});
}
else
{
break;
}
}
return res;
}
};

40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

All numbers (including target) will be positive integers.

The solution set must not contain duplicate combinations.

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,

A solution set is:

[

[1, 7],

[1, 2, 5],

[2, 6],

[1, 1, 6]

]

res.assign(sv.begin(),sv.end());//仅顺序容器定义了assign()，这里学会了这个成员的用法。注意array没有定义。

class Solution {
public:
void search(vector<int>& candidates,vector<vector<int>> &res,int i,int &target,int &sz,vector<int> v0)
{
for(int j=i;j<sz;++j)
{
int m=target-candidates[j];
if(target>candidates[j])
{
if(m>=candidates[i])
{
vector<int> v1{v0};
v1.push_back(candidates[j]);
search(candidates,res,j+1,m,sz,v1);
}
}
else if(!m)
{
vector<int> v1{v0};
v1.push_back(candidates[j]);
res.push_back(v1);//出现能够符合条件的加入结果
return;
}
else
{
return;
}
}
}

vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> res;
sort(candidates.begin(),candidates.end());//可以及时停止循环
int sz=candidates.size();
for(int i=0;i<sz;++i)
{
int m=target-candidates[i];
if(target>candidates[i])
{
if(m>=candidates[i])
{
search(candidates,res,i+1,m,sz,{candidates[i]});
}
}
else if(!m)
{
res.push_back({candidates[i]});
}
else
{
break;
}
}
set<vector<int>> sv(res.begin(),res.end());//去重
res.clear();
res.assign(sv.begin(),sv.end());//仅顺序容器定义了assign()
return res;
}
};

216. Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

class Solution {
public:
void search(vector<vector<int>> &res,int beg,int k,int n,vector<int> v)
{
if((!k)&&(!n))
{
res.push_back(v);
return;
}
for(int i=beg+1;i<=9&&i<=n;++i)
{
vector<int> v1{v};
v1.push_back(i);
search(res,i,k-1,n-i,v1);
}
}
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>> res;
vector<int> v{};
search(res,0,k,n,v);
return res;
}
};

377. Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]

target = 4

The possible combination ways are:

(1, 1, 1, 1)

(1, 1, 2)

(1, 2, 1)

(1, 3)

(2, 1, 1)

(2, 2)

(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.

Follow up:

What if negative numbers are allowed in the given array?

How does it change the problem?

What limitation we need to add to the question to allow negative numbers?

回溯超时，对我而言里面最难的一个，动态规划，其实我想到了利用target以前数的状态，但是还是不知道如何写，后来参考别人代码写出。

https://discuss.leetcode.com/topic/52186/my-3ms-java-dp-solution

class Solution {
public:
// int search(vector<int>& nums,int sz, int target)
// {
//     if(!target) return 1;
//     if(target<0)    return 0;
//     int res{0};
//     for(int i=0;i<sz&&nums[i]<=target;++i)
//     {
//         res+=search(nums,sz,target-nums[i]);
//     }
//     return  res;
// }

int combinationSum4(vector<int>& nums, int target) {
sort(nums.begin(),nums.end());
vector<int> res(target+1,0);
for(int i=1;i<=target;++i)
{
for(const int &im:nums)
{
if(im>i)
{
break;
}
else if(i==im)
{
res[i]+=1;//可能在此之前就已经有值
}
else
{
res[i]+=res[i-im];//利用前面已求状态得到现在状态。
}
}
}
return res[target];
}
};