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);}}}