DFS经典组合问题

下面将遇到的可以用递归求解的问题归纳于此

1. Combination

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]

【思路】递归

1.用一临时数组solution存储元素, 其大小应为k

2 base case , solution长度为k，并弹出最后元素

3 else 继续向solution push元素

注意：需要引用传参

 void getCombine(
                 int n,
                 int k,
                 int level,
                 vector<int> &solution,
                 vectorint>> &result){
     if(solution.size() == k){
         result.push_back(solution);
         return;
     }
     for(int i = level; i<=n; i++){
         solution.push_back(i);
         getCombine(n,k,i+1,solution,result);
         solution.pop_back();
     }
 }
 vectorint> > combine(int n, int k) {
     vector<int> solution;
     vectorint>> result;
     getCombine(n,k,1,solution,result);
     return result;
 }

2. 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]

【思路】 递归， 类似于上一题

base case: solution中元素之和sum为target

if sum值大于target 直接返回，并弹出最后元素

else 从candidate numbers 中继续选出元素加到solution中，同时用sum记录当前和值

注意：

1.结果中元素应保持non-descending order, 应先排序

2. 结果可以包含重复元素，递归中直接传level值就可以了，和上一题的不同. 

 void getCombineSum(
                    int &sum,
                    int level,
                    int target,
                    vector<int> &solution,
                    vector<int> &candidates,
                    vectorint>> &result
                    ){
     if(sum > target) return;
     if(sum == target){
         result.push_back(solution);
         return;
     }
     for(int i = level; i
         sum += candidates[i];
         solution.push_back(candidates[i]);
         getCombineSum(sum,"color:#ff6666;">i,target,solution,candidates,result);
         solution.pop_back();
         sum -= candidates[i];
     }

 }
 vectorint> > combinationSum(vector<int> &candidates, int target) {
     vector<int> solution;
     vectorint>> result;
     int sum = 0;
     sort(candidates.begin(),candidates.end());
     getCombineSum(sum,0,target,solution,candidates,result);
     return result;
 }

3.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] 

【思路】 与上一题类似，不同之处在于不能重复添加candidated中的元素

 void getCombineSum2(
                    int &sum,
                    int level,
                    int target,
                    vector<int> &solution,
                    vector<int> &num,
                    vectorint>> &result
                    ){
     if(sum > target) return;
     if(sum == target) {
         result.push_back(solution);
         return;
     }
     for(int i=level; i
         sum += num[i];
         solution.push_back(num[i]);
         getCombineSum2(sum,"color:#ff6666;">i+1,target,solution,num,result);
         sum -= num[i];
         solution.pop_back();
         "color:#ff6666;">while(i
     }
 }
 vectorint> > combinationSum2(vector<int> &num, int target) {
     vector<int> solution;
     vectorint>> result;
     int sum = 0;
     sort(num.begin(), num.end());
     getCombineSum2(sum,0,target,solution,num,result);
     return result;
 }