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一天一道LeetCode系列 ##（一）题目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]
##（二）解题
/*
主要思路：
1.对目标vector排序，定义一个临时的vector容器tmp
2.采用动态规划和回溯法的方法，对于当前要添加到tmp的数num
(1)如果target<num，return；
(2)如果target>num,则继续在num和num以后的数中选择一个数相加；(此处需要考虑可以重复的组合)
(3)如果target==num，则代表找到
对于(2),(3)递归完成后需要将压入的数弹出，即采用回溯法的思想
*/
class Solution {
public:
vector<vector<int>> ret;//结果
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(),candidates.end());//首先进行排序
for(int idx = 0;idx<candidates.size();idx++)
{
if(idx-1>=0 && candidates[idx] == candidates[idx-1]) continue;//避免重复查找
else{
if(candidates[idx]<=target){//如果小于则调用动态规划函数
vector<int> tmp;
tmp.push_back(candidates[idx]);
combinationDfs(candidates,tmp,idx,target-candidates[idx]);
}
}
}
return ret;
}
/*candidates为目标vector
tmp为临时vector变量，存储找到的组合
start是初始序号，代表在start及其以后的数字中查找
target是当前需要在candidates中查找的数
*/
void combinationDfs(vector<int>& candidates , vector<int>& tmp , int start,int target)
{
if(target == 0){//如果target等于0，代表第一个数就满足
ret.push_back(tmp);
return;
}
for(int idx = start;idx<candidates.size();idx++)
{
if(target > candidates[idx])
{
tmp.push_back(candidates[idx]);
combinationDfs(candidates,tmp,idx,target-candidates[idx]);
tmp.pop_back();//回溯法，要pop出最后压入的元素
}
else if(target == candidates[idx])//等于target代表以及找到
{
tmp.push_back(candidates[idx]);
ret.push_back(tmp);
tmp.pop_back();//回溯法，要pop出最后压入的元素
}
else if(target < candidates[idx])
{
return;
}
}
}
};