【Leetcode】Combination Sum II (Backtracking)

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] 

这道题和I有两点不同，第一是因为每个元素只能用一次，所以必须在递归处用i+1，第二个不同是在找重复的地方，如果一个数组{2，2，2}，第一次我用了三个2，第二次当我只能下{2}的时候，第二个元素应该添加第3个2，但是{2，2}的情况已经被考虑了，所以第3个2应该被跳过。

代码如下

public static ArrayList<ArrayList<Integer>> combinationSum2(int[] num,int target) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();ArrayList<Integer> temp = new ArrayList<Integer>();if (num == null || num.length == 0)return result;Arrays.sort(num);helper(num, 0, target, temp, result);return result;}public static void helper(int[] num, int start, int target,ArrayList<Integer> temp, ArrayList<ArrayList<Integer>> result) {if (target == 0) {result.add(new ArrayList<Integer>(temp));return;}if (target < 0)return;for (int i = start; i < num.length; i++) {if (i > start && num[i] == num[i - 1])continue;temp.add(num[i]);helper(num, i + 1, target - num[i], temp, result);temp.remove(temp.size() - 1);}}