1. 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.
The solution set must not contain duplicate combinations.
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
2. 思路
和上一题相比主要是两点,第一同一个数字不能重复,第二是当有重复候选时,同一个组合算一种。
直接复用上一次的代码,区别有两点:
递归时不是从当前位置开始限制,而是从下一个位置开始限制,也就是不可以使用当前这个已经使用过了的数字
每次在取值上对同一个targe的当前场景不和上一个重复。
3. 代码
耗时:23ms
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
return combinationSum(candidates, target, 0);
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target, int beg) {
vector<vector<int>> ret;
if (target <= 0) {return ret;}
for (int i = beg; i < candidates.size(); i++) {
if (i != beg && candidates[i] == candidates[i-1]) {continue;} // 避免重复
int iv = candidates[i];
if (iv == target) {
vector<int> one;
one.push_back(iv);
ret.push_back(one);
}
vector<vector<int>> suf = combinationSum(candidates, target-iv, i + 1);
for (int j = 0; j < suf.size(); j++) {
suf[j].push_back(iv);
ret.push_back(suf[j]);
}
}
return ret;
}
};
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