【leetcode】Array——Combination Sum I/II/III

Combination Sum

題目：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.
A solution set is: 
[7] 
[2, 2, 3] 

The same repeated number may be chosen from C unlimited number of times.

For example, given candidate set 2,3,6,7 and target 7, 

思路：經典的回溯

代碼：

public class Combination_Sum {
	public List<List<Integer>> combinationSum(int[] candidates, int target) {
		Arrays.sort(candidates);
		List<List<Integer>> results = new ArrayList<List<Integer>>();
		getResult(results,new ArrayList<Integer>(),candidates,0,target);

		return results;
	}
	private void getResult(List<List<Integer>> results,List<Integer>cur,int[]candidates,int start,int targer){
		if(targer>0){
			for(int i=start;i<candidates.length&&candidates[i]<=targer;i++){
				cur.add(candidates[i]);
				getResult(results, cur, candidates, i, targer-candidates[i]);
				cur.remove(cur.size()-1);}
		}else if(targer==0){
			results.add(new ArrayList<Integer>(cur));
		}

	}
}

Combination Sum II

題目：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.

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：和第一題思路一樣，但是回溯的時候，start參數要+1。此外，這道題還得考慮過濾重複的結果！

代碼：

public class Combination_Sum_II {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    	Arrays.sort(candidates);
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        List<Integer> cur = new ArrayList<Integer>();
        getResult(results,cur, candidates, 0, target);
        return results;
    }
    private void getResult(List<List<Integer>> results, List<Integer> cur,int[]candidates,int start,int target){
    	if(target==0){
    		results.add(new ArrayList<>(cur));
    		return;
    	}
    	for(int i=start;i<candidates.length&&candidates[i]<=target;i++){
    		//過濾重複的結果！i>start這個判斷很關鍵
    		if(i>=1&&candidates[i]==candidates[i-1]&&i>start) continue;
    		cur.add(candidates[i]);
    		getResult(results, cur, candidates, i+1, target-candidates[i]);
    		cur.remove(cur.size()-1);
    	}
    }
}

Combination Sum III

題目：

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]]

思路：仍然使用回溯。但是要經行當前List長度判斷，>=k的時候不能在遞歸了，而是直接return，然後remove

代碼：

public class Combination_Sum_III {
	public List<List<Integer>> combinationSum3(int k, int n) {
		List<List<Integer>> results = new ArrayList<List<Integer>>();
		List<Integer>cur = new ArrayList<Integer>();
		getResult(results, cur, 1, n,k);

		return results;
	}

	private void getResult(List<List<Integer>> results,List<Integer>cur,int start,int n,int k){
		if(n==0&&cur.size()==k){
			results.add(new ArrayList<Integer>(cur));
		}else if(n>0&&cur.size()<k){
			for(int i=start;i<=9;i++){
				cur.add(i);
				getResult(results, cur, i+1, n-i, k);
				cur.remove(cur.size()-1);
			}
		}
	}
}