2014
11-18

# LeetCode-Combination Sum II[DFS]

### Combination Sum II

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, a1a2 ≤ … ≤ ak).
• 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]

// LeetCode, 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> intermediate;
dfs(nums, target, 0, intermediate, result);
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(vector<int> &nums, int gap, int start,
vector<int> &intermediate, vector<vector<int> > &result) {
if (gap == 0) {  //  找到一个合法解
result.push_back(intermediate);
return;
}

int previous = -1;
for (size_t i = start; i < nums.size(); i++) {
// 如果上一轮循环没有选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];

intermediate.push_back(nums[i]);
dfs(nums, gap - nums[i], i + 1, intermediate, result);
intermediate.pop_back();  // 恢复环境
}
}
};


Combination Sum

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2. 第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。