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]
简单的回溯法(递归实现).
比如对于数组3,2,6,7,target = 7,对数组排序得到[2,3,6,7]
1、第1个数字选取2, 那么接下来就是解决从数组[2,3,6,7]选择数字且target = 7-2 = 5
2、第2个数字选择2，那么接下来就是解决从数组[2,3,6,7]选择数字且target = 5-2 = 3
3、第3个数字选择2，那么接下来就是解决从数组[2,3,6,7]选择数字且target = 3-2 = 1
4、此时target = 1小于数组中的所有数字，失败，回溯，重新选择第3个数字
5、第3个数字选择3，那么接下来就是解决从数组[2,3,6,7]选择数字且target = 3-3 = 0
6、target = 0，找到了一组解，继续回溯寻找其他解
需要注意的是：如果数组中包含重复元素，我们要忽略（因为每个数字可以选择多次，如果不忽略的话，就会产生重复的结果）。貌似oj的测试集数组中都不包含重复的数字
class Solution {
private:
vector<vector<int> > res;
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
sort(candidates.begin(), candidates.end());//为了输出结果递增，因此先对数组排序
vector<int> tmpres;
helper(candidates, 0, target, tmpres);
return res;
}
//从数组candidates[index,...]寻找和为target的组合
void helper(vector<int> &candidates, const int index, const int target, vector<int>&tmpres)
{
if(target == 0)
{
res.push_back(tmpres);
return;
}
for(int i = index; i < candidates.size() && target >= candidates[i]; i++)
if(i == 0 || candidates[i] != candidates[i-1])//由于每个数可以选取多次，因此数组中重复的数就不用考虑
{
tmpres.push_back(candidates[i]);
helper(candidates, i, target - candidates[i], tmpres);
tmpres.pop_back();
}
}
};
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]
和上一题差不多，只是每个元素只能选一次。
由于有重复元素的存在，比如数组为[1(1),1(2),2,3]，target = 6. 可能出现重复结果1(1),2,3 和 1(2),2,3 本文地址
我们可以如下处理：如果数组中当前的数字出现重复，在前面重复了k次，且临时结果数组中也包含了k个当前数字，那么当前的数字可以选择；否则就不选择当前数字
class Solution {
private:
vector<vector<int> >res;
public:
vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<int> tmpres;
helper(candidates, 0, target, tmpres, 0);
return res;
}
//从数组candidates[index,...]寻找和为target的组合,times为前一个数字candidates[index-1]重复出现的次数
void helper(vector<int> &candidates, const int index, const int target, vector<int>&tmpres, int times)
{
if(target == 0)
{
res.push_back(tmpres);
return;
}
for(int i = index; i < candidates.size() && target >= candidates[i]; i++)
{
if(i > 0 && candidates[i] == candidates[i-1])times++;
else times = 1;
if(times == 1 || (tmpres.size() >= times-1 && tmpres[tmpres.size()-times+1] == candidates[i]))
{
tmpres.push_back(candidates[i]);
helper(candidates, i+1, target - candidates[i], tmpres, times);
tmpres.pop_back();
}
}
}
};
还有一种方法是，在每个子问题的数组中，重复的数字都不选择，这种更简洁
class Solution {
private:
vector<vector<int> >res;
public:
vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<int> tmpres;
helper(candidates, 0, target, tmpres);
return res;
}
//从数组candidates[index,...]寻找和为target的组合
void helper(vector<int> &candidates, const int index, const int target, vector<int>&tmpres)
{
if(target == 0)
{
res.push_back(tmpres);
return;
}
for(int i = index; i < candidates.size() && target >= candidates[i]; i++)
{
if(i > index && candidates[i] == candidates[i-1])continue;//当前子问题中，重复数字都不选择
tmpres.push_back(candidates[i]);
helper(candidates, i+1, target - candidates[i], tmpres);
tmpres.pop_back();
}
}
};
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