LeetCode——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]

原题链接：https://oj.leetcode.com/problems/combination-sum-ii/

与之前一题的区别在于每个数字只能使用一次。故需在递归的时候直接使用下一个数字。

public class CombinationSumII {
	public static void main(String[] args) {
		List<List<Integer>> list = new CombinationSumII().combinationSum2(new int[]{10,1,2,7,6,1,5},8);
		for(List<Integer> li : list){
			for(Integer i : li)
				System.out.print(i + "\t");
			System.out.println();
		}
	}
    private List<Integer> list = new  ArrayList<Integer>();
	private List<List<Integer>> result = new ArrayList<List<Integer>>();
	public List<List<Integer>> combinationSum2(int[] num, int target) {
		if(num.length == 0)
			return result;
		Arrays.sort(num);
		dfs(num,target,0);
		return result;
	}
	public void dfs(int[] candidates,int target,int index){
		if(target < 0)
			return;
		if(target == 0){
			result.add(new ArrayList<Integer>(list));
			return;
		}
		for(int i=index;i<candidates.length;i++){
			if(i>index && candidates[i] == candidates[i-1])
				continue;
			list.add(candidates[i]);
			dfs(candidates,target-candidates[i],i+1);
			list.remove(list.size()-1);
		}
	}
}