LeetCode—数组（3）

1.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] 

 public List<List<Integer>> combinationSum(int[] candidates, int target) {         List<List<Integer>> res=new ArrayList<List<Integer>>();      if(candidates==null||candidates.length==0)        return res;                Arrays.sort(candidates);        combinationSum(candidates, target,0,new ArrayList<Integer>(),res);        return res;           }    private static void combinationSum(int[] candidates, int target, int start, ArrayList<Integer> list,List<List<Integer>> res) {if(target==0){res.add(list);return;}for(int i=start;i<candidates.length;i++){if(target-candidates[i]>=0){ArrayList<Integer> tmp=(ArrayList<Integer>)list.clone();int tmp_target=target-candidates[i];tmp.add(candidates[i]);combinationSum(candidates, tmp_target, i, tmp, res);}}}</span>

2.

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] 

public List<List<Integer>> combinationSum2(int[] candidates, int target) {          List<List<Integer>> res=new ArrayList<List<Integer>>();        if(candidates==null||candidates.length==0)        return res;        Arrays.sort(candidates);        combinationSum2(candidates,target,0,new ArrayList<Integer>(),res);        return res;    }private static void combinationSum2(int[] candidates, int target, int start, ArrayList<Integer> arrayList,List<List<Integer>> res) {if(target==0){res.add(arrayList);return;}for(int i=start;i<candidates.length;i++){if (i > start && candidates[i] == candidates[i-1]) continue;if(target-candidates[i]>=0){ArrayList<Integer> tmp=(ArrayList<Integer>)arrayList.clone();int tmp_target=target-candidates[i];tmp.add(candidates[i]);combinationSum2(candidates, tmp_target, i+1, tmp, res);}}}</span>

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.

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]]

public List<List<Integer>> combinationSum3(int k, int n) {        List<List<Integer>> res = new ArrayList<>();        Deque<Integer> tmp = new ArrayDeque<>();        if (n == 0 || k == 0 || n / k > 9) {            return res;        }        helper(res, tmp, 1, k, n);        return res;    }    private void helper(List<List<Integer>> res, Deque<Integer> tmp, int start, int k, int n) {        if (0 == n && k == 0) {            res.add(new ArrayList<>(tmp));            return;        }        for (int i = start; i <= 9; i++) {            tmp.addLast(i);            helper(res, tmp, i + 1, k - 1, n - i);            tmp.removeLast();        }    }</span>