1. 每个元素可用多次

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] 

void cal(vector
     
      & candidates, int start, int target, vector
      
       & v, vector
       
        
         >& ans){    if(0 == target)    {        ans.push_back(v);        return;    }    if(target < 0)        return;    int n = candidates.size(), i;    for(i = start; i < n; i++)    {        if(i > start && candidates[i] == candidates[i-1])            continue;        v.push_back(candidates[i]);        cal(candidates, i, target-candidates[i], v, ans);        v.pop_back();    }}vector
         
          
           > combinationSum(vector
           
            & candidates, int target) { vector
            
             
              > ans; vector
              
                v; sort(candidates.begin(), candidates.end()); cal(candidates, 0, target, v, ans); return ans;}
              
             
            
           
          
         
        
       
      
     

 

2. 每个元素只能用一次

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] 

void cal(vector
     
      & candidates, int start, int target, vector
      
       & v, vector
       
        
         >& ans){    if(0 == target)    {        ans.push_back(v);        return;    }    if(target < 0)        return;    int n = candidates.size(), i;    for(i = start; i < n; i++)    {        if(i > start && candidates[i] == candidates[i-1])            continue;        v.push_back(candidates[i]);        cal(candidates, i+1, target-candidates[i], v, ans);        v.pop_back();    }}vector
         
          
           > combinationSum2(vector
           
            & candidates, int target) { vector
            
             
              > ans; vector
              
                v; sort(candidates.begin(), candidates.end()); cal(candidates, 0, target, v, ans); return ans;}
              
             
            
           
          
         
        
       
      
     

if(i > start && candidates[i] == candidates[i-1])continue;中的i > start可以保证没有重复结果。

 

3. 数字范围1-9，给定个数。

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

 [[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

void cal(vector
      
       & candidates, int start, int target, int k, vector
       
        & v, vector
        
         
          >& ans){    if(0 == target && 0 == k)    {        ans.push_back(v);        return;    }    if(target < 0 || 0 == k)        return;    int n = candidates.size(), i;    for(i = start; i < n; i++)    {        if(i > start && candidates[i] == candidates[i-1])            continue;        v.push_back(candidates[i]);        cal(candidates, i+1, target-candidates[i], k-1, v, ans);        v.pop_back();    }}vector
          
           
            > combinationSum3(int k, int target) { vector
            
             
              > ans; vector
              
                candidates, v; for(int i = 1; i <= 9; i++) candidates.push_back(i); cal(candidates, 0, target, k, v, ans); return ans;}