LeetCode 题解（175): Combination Sum II
题目：
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.
10,1,2,7,6,1,5 and target
8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
题解：
与Combination Sum I 的区别是元素不能重复使用，结果中不能有重复结果。所以要跳过排序后重复的元素。
C++版：
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> results;
sort(candidates.begin(), candidates.end());
int i = 0;
while(i < candidates.size()) {
vector<int> result;
result.push_back(candidates[i]);
combination(candidates, results, result, target - candidates[i], i + 1);
result.pop_back();
while(candidates[i+1] == candidates[i])
i++;
i++;
}
return results;
}
void combination(vector<int>& candidates, vector<vector<int>>& results, vector<int>& result, int target, int i) {
if(target < 0)
return;
if(target == 0) {
results.push_back(result);
return;
}
for(int j = i; j < candidates.size(); j++) {
result.push_back(candidates[j]);
combination(candidates, results, result, target - candidates[j], j + 1);
result.pop_back();
while(candidates[j] == candidates[j+1])
j++;
}
}
};
Java版：
public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> results = new ArrayList<List<Integer>>();
for(int i = 0; i < candidates.length; i++) {
List<Integer> result = new ArrayList<>();
result.add(candidates[i]);
combination(candidates, results, result, target - candidates[i], i + 1);
result.remove(result.size() - 1);
while(i + 1 < candidates.length && candidates[i] == candidates[i + 1])
i++;
}
return results;
}
public void combination(int[] candidates, List<List<Integer>> results, List<Integer> result, int target, int i) {
if(target < 0)
return;
if(target == 0) {
List<Integer> temp = new ArrayList<>();
temp.addAll(result);
results.add(temp);
}
for(int j = i; j < candidates.length; j++) {
result.add(candidates[j]);
combination(candidates, results, result, target - candidates[j], j + 1);
result.remove(result.size() - 1);
while(j + 1 < candidates.length && candidates[j] == candidates[j + 1])
j++;
}
}
}
Python版：
import copy
class Solution:
# @param {integer[]} candidates
# @param {integer} target
# @return {integer[][]}
def combinationSum2(self, candidates, target):
results, result = [], []
candidates.sort()
i = 0
while i < len(candidates):
result.append(candidates[i])
self.combination(candidates, results, result, target - candidates[i], i + 1)
result.pop()
while i + 1 < len(candidates) and candidates[i] == candidates[i+1]:
i += 1
i += 1
return results
def combination(self, candidates, results, result, target, i):
if target < 0:
return
if target == 0:
temp = copy.copy(result)
results.append(temp)
j = i
while j < len(candidates):
result.append(candidates[j])
self.combination(candidates, results, result, target - candidates[j], j + 1)
result.pop()
while j + 1 < len(candidates) and candidates[j] == candidates[j+1]:
j += 1
j += 1