Combination Sum II
描述
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 () must be in non-descending order. (ie, ).
- 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]
分析
无
代码
// Combination Sum II
// 时间复杂度O(n!),空间复杂度O(n)
class Solution {
public:
vector<vector<int> > combinationSum2(vector<int> &nums, int target) {
sort(nums.begin(), nums.end()); // 跟第 50 行配合,
// 确保每个元素最多只用一次
vector<vector<int> > result;
vector<int> path;
dfs(nums, path, result, target, 0);
return result;
}
private:
// 使用nums[start, nums.size())之间的元素,能找到的所有可行解
static void dfs(const vector<int> &nums, vector<int> &path,
vector<vector<int> > &result, int gap, int start) {
if (gap == 0) { // 找到一个合法解
result.push_back(path);
return;
}
int previous = -1;
for (size_t i = start; i < nums.size(); i++) {
// 如果上一轮循环已经使用了nums[i],则本次循环就不能再选nums[i],
// 确保nums[i]最多只用一次
if (previous == nums[i]) continue;
if (gap < nums[i]) return; // 剪枝
previous = nums[i];
path.push_back(nums[i]);
dfs(nums, path, result, gap - nums[i], i + 1);
path.pop_back(); // 恢复环境
}
}
};