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]
]
39. Combination Sum的变形，39题数组中的数字可以重复使用，而这道题数组中的数字不能重复使用。这里要考虑跳过重复的数字，其它的与39题一样。
解法：和39一样，递归 + backtracking
Java:
public List<List<Integer>> combinationSum2(int[] cand, int target) {
Arrays.sort(cand);
List<List<Integer>> res = new ArrayList<List<Integer>>();
List<Integer> path = new ArrayList<Integer>();
dfs_com(cand, 0, target, path, res);
return res;
}
void dfs_com(int[] cand, int cur, int target, List<Integer> path, List<List<Integer>> res) {
if (target == 0) {
res.add(new ArrayList(path));
return ;
}
if (target < 0) return;
for (int i = cur; i < cand.length; i++){
if (i > cur && cand[i] == cand[i-1]) continue;
path.add(path.size(), cand[i]);
dfs_com(cand, i+1, target - cand[i], path, res);
path.remove(path.size()-1);
}
}
Java:
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
List<Integer> curr = new ArrayList<Integer>();
Arrays.sort(candidates);
helper(result, curr, 0, target, candidates);
return result;
}
public void helper(List<List<Integer>> result, List<Integer> curr, int start, int target, int[] candidates){
if(target==0){
result.add(new ArrayList<Integer>(curr));
return;
}
if(target<0){
return;
}
int prev=-1;
for(int i=start; i<candidates.length; i++){
if(prev!=candidates[i]){ // each time start from different element
curr.add(candidates[i]);
helper(result, curr, i+1, target-candidates[i], candidates); // and use next element only
curr.remove(curr.size()-1);
prev=candidates[i];
}
}
}
Python:
class Solution:
# @param candidates, a list of integers
# @param target, integer
# @return a list of lists of integers
def combinationSum2(self, candidates, target):
result = []
self.combinationSumRecu(sorted(candidates), result, 0, [], target)
return result def combinationSumRecu(self, candidates, result, start, intermediate, target):
if target == 0:
result.append(list(intermediate))
prev = 0
while start < len(candidates) and candidates[start] <= target:
if prev != candidates[start]:
intermediate.append(candidates[start])
self.combinationSumRecu(candidates, result, start + 1, intermediate, target - candidates[start])
intermediate.pop()
prev = candidates[start]
start += 1
C++:
class Solution {
public:
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
vector<vector<int> > res;
vector<int> out;
sort(num.begin(), num.end());
combinationSum2DFS(num, target, 0, out, res);
return res;
}
void combinationSum2DFS(vector<int> &num, int target, int start, vector<int> &out, vector<vector<int> > &res) {
if (target < 0) return;
else if (target == 0) res.push_back(out);
else {
for (int i = start; i < num.size(); ++i) {
if (i > start && num[i] == num[i - 1]) continue;
out.push_back(num[i]);
combinationSum2DFS(num, target - num[i], i + 1, out, res);
out.pop_back();
}
}
}
};
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