Combination Sum （II）

题目：

1.Given a set of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums toT.

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, … ,a_k) must be in non-descending order. (ie,a₁ ≤a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target7,
A solution set is:
[7]

[2, 2, 3]




2.Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums toT.

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, … ,a_k) must be in non-descending order. (ie,a₁ ≤a₂ ≤ … ≤ a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

思想：

对于求和题，我们可以采用动态规划的思想来处理。例如：对于目标数7，如果我们已经已有的数为2，那么只要能够在数组中找到一个子序列，使得子序列之和为5即可。这样一来我们可以采用递归的方式来解题。题1与题2的不同点是，题1有序且数据可重复使用，题2无序不可重复使用。这就要求题1与题2的递归方式不同，且为需对题2的原数组进行排序操作。二者的不同之处，已在代码中用红色标记出。

代码1：

class Solution {public:vector<vector<int> > combinationSum(vector<int>& candidates, int target) {vector<vector<int> > ret;vector<int > combination;combinationSum(candidates, target, ret, combination, 0);return ret;}private:void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &ret, vector<int> &combination, int begin){if (!target)//已经搜索到一个答案{ret.push_back(combination);return;}for (int  i = begin; i < candidates.size()&& target>=candidates[i]; i++){combination.push_back(candidates[i]);combinationSum(candidates, target - candidates[i], ret, combination, i);//递归搜索combination.pop_back();//若该集合不满足条件，则弹出}}};

代码2：

class Solution {public:vector<vector<int> > combinationSum(vector<int>& candidates, int target) {vector<vector<int> > ret;vector<int > combination;<span style="color:#FF0000;">sort(candidates.begin(), candidates.end());</span>combinationSum(candidates, target, ret, combination, 0);return ret;}private:void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &ret, vector<int> &combination, int begin){if (!target)//已经搜索到一个答案{ret.push_back(combination);return;}for (int i = begin; i < candidates.size() && target >= candidates[i]; i++){<span style="color:#FF0000;">if (i == begin || candidates[i] != candidates[i - 1])</span>{//巧妙的避免了，当i=0时，i-1会出现溢出的现象combination.push_back(candidates[i]);combinationSum(candidates, target - candidates[i], ret, combination, <span style="color:#FF0000;">i+1</span>);//递归搜索combination.pop_back();//若该集合不满足条件，则弹出}}}};