40 Combination Sum II 组合总和 II
Description:
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:
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]
]
题目描述:
给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用一次。
说明:
所有数字（包括目标数）都是正整数。
解集不能包含重复的组合。
示例 :
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
所求解集为:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:
[
[1,2,2],
[5]
]
思路:
回溯法
- 先对 candidates排序, 保证数组从小到大排列
- 从 candidates中按大小顺序选出元素加入候选列表
- 如果 target == 0, 将候选列表加入到结果中
- 撤销选择
注意这里的元素不可重复选择, 所以选择下一层递归函数的时候要从下一个下标开始选取, 并且要去重
时间复杂度O(n!), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
vector> combinationSum2(vector& candidates, int target)
{
vector> result;
vector temp;
sort(candidates.begin(), candidates.end());
trackback(result, candidates, target, 0, temp);
return result;
}
void trackback(vector> &result, vector &candidates, int target, int start, vector &temp)
{
if (target == 0) result.push_back(temp);
for (int i = start; i < candidates.size(); i++)
{
if (i > start and candidates[i] == candidates[i - 1]) continue;
if (target < 0) break;
temp.push_back(candidates[i]);
trackback(result, candidates, target - candidates[i], i + 1, temp);
temp.pop_back();
}
}
};
Java:
class Solution {
public List> combinationSum2(int[] candidates, int target) {
List> result = new ArrayList<>();
Arrays.sort(candidates);
trackback(result, candidates, target, 0, new ArrayList<>());
return result;
}
private void trackback(List> result, int[] candidates, int target, int start, List list) {
if (target == 0) {
result.add(new ArrayList<>(list));
return;
}
for (int i = start; i < candidates.length; i++) {
if (i > start && candidates[i] == candidates[i - 1]) continue;
if (target < 0) break;
list.add(candidates[i]);
trackback(result, candidates, target - candidates[i], i + 1, list);
list.remove(list.size() - 1);
}
}
}
Python:
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
def helper(candidates: List[int], target: int) -> List[List[int]]:
if not target:
return [[]]
elif not candidates or target < min(candidates):
return []
return [[candidates[0]] + x for x in helper(candidates[1:], target - candidates[0])] + helper([i for i in candidates if i > candidates[0]], target)
return helper(sorted(candidates), target)