Combinations 集合

39. 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.
• 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]]

允许重复使用集合中给定的数字。

public class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) {    Arrays.sort(candidates);    List<List<Integer>> result=new ArrayList<List<Integer>>();        getResult(result,new ArrayList<Integer>(),candidates,target,0);    return result;    }        private void getResult(List<List<Integer>> result, List<Integer> cur, int candidates[], int target, int start){        if(target>0){            for(int i=start;i<candidates.length&&candidates[i]<=target;i++){                cur.add(candidates[i]);                getResult(result,cur,candidates,target-candidates[i],i); //体现在这里，如果上一次添加了candidates[0]这次还是从[0] 开始可以继续添加                cur.remove(cur.size()-1);            }        }        else if (target==0){            result.add(new ArrayList<Integer>(cur));        }            }}

40. 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.
• 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]]

不允许重复使用给定集合中的数字，但如果有两个1  则1可以使用两次 

在上述代码中做改变，在添加candidates[i]之后  从i+1开始寻找而不是从i开始 避免重复使用给定数字，在最后添加结果集时检查是否存在重复情况

public class Solution {    public List<List<Integer>> combinationSum2(int[] candidates, int target) {         Arrays.sort(candidates);    List<List<Integer>> result=new ArrayList<List<Integer>>();        getResult(result,new ArrayList<Integer>(),candidates,target,0);    return result;    }        private void getResult(List<List<Integer>> result, List<Integer> cur, int candidates[], int target, int start){        if(target>0){            for(int i=start;i<candidates.length&&candidates[i]<=target;i++){                cur.add(candidates[i]);                getResult(result,cur,candidates,target-candidates[i],i+1);                 cur.remove(cur.size()-1);            }        }        else if (target==0){            if(!result.contains(cur))            result.add(new ArrayList<Integer>(cur));        }            }}

216. Combination Sum III

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.

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]]

和上题类似，只不过candidates变成了1-9，结果集添加条件变为 k==0&&n==0 需要两者条件同时满足，不满足的情况应该是在两者不同时到达0，会有一个先小于0,

可以用条件n<0/n<0||k<0  来进行判断

注意在添加list时要 new一个ArrayList 来添加，list是引用类型  用后变空！！

public class Solution {    public List<List<Integer>> combinationSum3(int k, int n) {        List<List<Integer>> result=new ArrayList<List<Integer>>();        List<Integer> list=new ArrayList<Integer>();        getCombination(result,new ArrayList<Integer>(),1,n,k);        return result;    }    public void getCombination(List<List<Integer>> result,List<Integer> list,int start,int n,int k){        if(n<0) return ;        if(k==0&&n==0)  result.add(new ArrayList<Integer>(list));        for(int i=start;i<=9;i++){            list.add(i);            getCombination(result,list,i+1,n-i,k-1);            list.remove(list.size()-1);        }    }}

377. Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]target = 4The possible combination ways are:(1, 1, 1, 1)(1, 1, 2)(1, 2, 1)(1, 3)(2, 1, 1)(2, 2)(3, 1)Note that different sequences are counted as different combinations.Therefore the output is 7.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

用动态规划的做法：状态转移方程: combinationSum4(x): f(x)
f(target) = f(target - nums[0]) + f(target - nums[1]) + ... + f(target - nums[nums.length-1])

public class Solution {    public int combinationSum4(int[] nums, int target) {       int []dp=new int[target+1];       Arrays.sort(nums);       for(int i=0;i<dp.length;i++){           for(int j=0;j<nums.length;j++){               if(i<nums[j]) break;               if(i==nums[j]) dp[i]++;               else dp[i]+=dp[i-nums[j]];           }       }       return dp[target];    }}

仍旧用回溯的做法，超时

public class Solution {    public int combinationSum4(int[] nums, int target) {        List<List<Integer>> result=new ArrayList<List<Integer>>();        getCom(result,new ArrayList<Integer>(),nums,target);        return result.size();    }    public void getCom(List<List<Integer>> result,List<Integer> list,int []nums,int target){        if(target<0) return ;        if(target==0) result.add(new ArrayList(list));        for(int i=0;i<nums.length;i++){            list.add(nums[i]);            getCom(result,list,nums,target-nums[i]);            list.remove(list.size()-1);        }    }}

77. Combinations

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],]

给定范围和数字个数，返回它们能组成的符合要求的所有集合 

从1开始往后找没添加一个就将k-1 直到k=0 

回溯 递归

public class Solution {    public List<List<Integer>> combine(int n, int k) {        List<List<Integer>> combs=new ArrayList<List<Integer>>();        combine(combs,new ArrayList<Integer>(),1,n,k);        return combs;    }    public static void combine(List<List<Integer>> combs,List<Integer> comb,int start,int n,int k){        if(k==0){            combs.add(new ArrayList<Integer>(comb));            return;        }        for(int i=start;i<=n;i++){            comb.add(i);            combine(combs,comb,i+1,n,k-1);            comb.remove(comb.size()-1);        }    }}