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]
跟 Combination Sum 不同的是，候选数集中的元素只能用一次，且给定的候选数集中的元素会存在重复，这样我们对其排序尽管可以保证解元素不减有序，但重复的问题就避免不了，因为我们不知道是重复的数构成的解，每个元素我们都是相同对待的。如例子中给定的，我们可以有前面的1跟 7 够成解，其实后面的 1 跟 7 也构成的了解，就要求对最终的结果去重。Combination Sum 的代码稍作修改就可以了。
class Solution {
public:
/*
cur_sum: 当前的累加和
icur : 当前候选数的下标
target ：目标值
cur_re : 当前的潜在解
re : 最终的结果集
can : 候选数集
*/
void dfs(int cur_sum, int icur, int target, vector<int>& cur_re, vector<vector<int> > &re, vector<int> can)
{
if(cur_sum > target || icur > can.size()) return;
if(cur_sum == target){
re.push_back(cur_re);
return;
}
//(target - cur_sum >= can[i]类似剪枝，加上该判定与
//不加，时间差4-5倍。
for(int i = icur; i < can.size() && (target - cur_sum >= can[i]); ++i){
cur_sum += can[i];
cur_re.push_back(can[i]);
//each item only be used once, so handle the next one.
dfs(cur_sum, i + 1, target, cur_re, re, can);
//go back
cur_re.pop_back();
cur_sum -= can[i];
}
}
//why not use set?
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
vector<vector<int> > re;
vector<int> cur_re;
if(num.size() == 0) return re;
sort(num.begin(), num.end());
if(num[0] > target) return re;
dfs(0, 0, target, cur_re, re, num);
//erase the duplicate
sort(re.begin(),re.end());
vector<vector<int> >::iterator vvit;
vvit=unique(re.begin(),re.end());
re.erase(vvit,re.end());
return re;
}
};