40. Combination Sum II
給定一組候選數字（candidates）和一個目標數字（target），找到候選數字中所有和為目標數字的唯一組合。
在每個組合中，候選數字中的每個數字只能使用一次。
注意：解集不能包含重複的組合。
範例 1：
輸入：candidates = [10,1,2,7,6,1,5], target = 8
輸出：
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
Python
from typing import List
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
candidates.sort()
result = []
def backtrack(start, path, remaining):
if remaining == 0:
result.append(list(path))
return
if remaining < 0:
return
for i in range(start, len(candidates)):
if i > start and candidates[i] == candidates[i - 1]:
continue
path.append(candidates[i])
backtrack(i + 1, path, remaining - candidates[i])
path.pop()
backtrack(0, [], target)
return result
16.60MB, 70ms
C++
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> result;
vector<int> combination;
sort(candidates.begin(), candidates.end());
backtrack(candidates, target, 0, combination, result);
return result;
}
private:
void backtrack(vector<int>& candidates, int target, int start, vector<int>&
combination, vector<vector<int>>& result) {
if (target == 0) {
result.push_back(combination);
return;
}
if (target < 0) {
return;
}
for (int i = start; i < candidates.size(); ++i) {
if (i > start && candidates[i] == candidates[i - 1]) {
continue; // skip duplicates
}
combination.push_back(candidates[i]);
backtrack(candidates, target - candidates[i], i + 1, combination, result);
combination.pop_back();
}
}
};
12.53MB, 6ms
Javascript
/**
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/
var combinationSum2 = function(candidates, target) {
const result = [];
candidates.sort((a, b) => a - b);
const backtrack = (start, path, remaining) => {
if (remaining === 0) {
result.push([...path]);
return;
}
if (remaining < 0) {
return;
}
for (let i = start; i < candidates.length; i++) {
if (i > start && candidates[i] === candidates[i - 1]) {
continue; // Skip duplicates
}
path.push(candidates[i]);
backtrack(i + 1, path, remaining - candidates[i]);
path.pop();
}
};
backtrack(0, [], target);
return result;
};
53.27MB, 64ms
