題目:
40. Combination Sum II
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates 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.
Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]
思路:
遞歸
代碼:
class Solution {
public:
void help(vector<int>& nums, vector<vector<int> >& vec, vector<int>& v, int start, int target){
if(target < 0)return;
if(target == 0){
vec.push_back(v);
return;
}
for(int i = start; i < nums.size(); ++i){
//跳過同一層的重複數值
if(i > start && nums[i] == nums[i-1])continue;
v.push_back(nums[i]);
help(nums, vec, v, i+1, target-nums[i]);
v.pop_back();
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int> > vec;
vector<int> v;
if(candidates.size() < 1)return vec;
sort(candidates.begin(), candidates.end());
help(candidates, vec, v, 0, target);
return vec;
}
};