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Leetcode:Combination Sum 子集和问题
Combination Sum：
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7] [2, 2, 3]
解题分析：
class Solution {public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<vector<int>> result; if (candidates.size() == 0) return result; sort(candidates.begin(), candidates.end()); // sort vector<int> path; dfs(candidates, target, 0, path, result); return result; } void dfs(const vector<int>& nums, int gap, int start, vector<int>& path, vector<vector<int>>& result) { if (gap == 0) { // 找到一个合法解 result.push_back(path); return; } for (int i = start; i < nums.size(); ++i) { // 扩展状态 if (gap < nums.at(i)) continue; // 剪枝 path.push_back(nums.at(i)); // 执行扩展动作 dfs(nums, gap - nums.at(i), i, path, result); path.pop_back(); // 撤销扩展动作 } }};
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, a1 ≤ a2 ≤ … ≤ 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]
class Solution {public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int>> result; if (num.size() == 0) return result; sort(num.begin(), num.end()); vector<int> path; dfs(num, target, 0, path, result); return result; } void dfs(const vector<int>& num, int gap, int start, vector<int>& path, vector<vector<int>>& result) { if (gap == 0) { // 找到一个合法解 result.push_back(path); return; } int prev = -1; for (int i = start; i < num.size(); ++i) { if (prev == num.at(i)) continue; // 确保num.at(i) 最多只用一次 if (gap < num.at(i)) return; // 剪枝 prev = num.at(i); path.push_back(num.at(i)); // 执行扩展动作 dfs(num, gap - num.at(i), i + 1, path, result); path.pop_back(); // 撤消扩展动作 } }};
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