39. Combination Sum

1. 每个元素可用多次

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.

Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

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]

void cal(vector<int>& candidates, int start, int target, vector<int>& v, vector<vector<int>>& ans)
{
if(0 == target)
{
ans.push_back(v);
return;
}
if(target < 0)
return;
int n = candidates.size(), i;
for(i = start; i < n; i++)
{
if(i > start && candidates[i] == candidates[i-1])
continue;
v.push_back(candidates[i]);
cal(candidates, i, target-candidates[i], v, ans);
v.pop_back();
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> ans;
vector<int> v;
sort(candidates.begin(), candidates.end());
cal(candidates, 0, target, v, ans);
return ans;
}

2. 每个元素只能用一次

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.

Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

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]

void cal(vector<int>& candidates, int start, int target, vector<int>& v, vector<vector<int>>& ans)
{
if(0 == target)
{
ans.push_back(v);
return;
}
if(target < 0)
return;
int n = candidates.size(), i;
for(i = start; i < n; i++)
{
if(i > start && candidates[i] == candidates[i-1])
continue;
v.push_back(candidates[i]);
cal(candidates, i+1, target-candidates[i], v, ans);
v.pop_back();
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> ans;
vector<int> v;
sort(candidates.begin(), candidates.end());
cal(candidates, 0, target, v, ans);
return ans;
}

if(i > start && candidates[i] == candidates[i-1])continue;中的i > start可以保证没有重复结果。

3. 数字范围1-9，给定个数。

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.

Ensure that numbers within the set are sorted in ascending order.

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]]

void cal(vector<int>& candidates, int start, int target, int k, vector<int>& v, vector<vector<int>>& ans)
{
if(0 == target && 0 == k)
{
ans.push_back(v);
return;
}
if(target < 0 || 0 == k)
return;
int n = candidates.size(), i;
for(i = start; i < n; i++)
{
if(i > start && candidates[i] == candidates[i-1])
continue;
v.push_back(candidates[i]);
cal(candidates, i+1, target-candidates[i], k-1, v, ans);
v.pop_back();
}
}
vector<vector<int>> combinationSum3(int k, int target) {
vector<vector<int>> ans;
vector<int> candidates, v;
for(int i = 1; i <= 9; i++)
candidates.push_back(i);
cal(candidates, 0, target, k, v, ans);
return ans;
}