40. 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.
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]]

思路：
既然数组中存在重复的值，就要注意可能会将重复的情况加入结果数组。例如，如果数组为[2,2,2,6],目标值为8，可能会在结果数组中出现多次[2,6]。
同样的，在进行递归的子遍历的时候也要注意，可能会出现重复值，例如数组为[2,2,2,6],目标值为10，则结果集[2,2,6]也可能出现多次，所以在子遍历中也要记得去除重复情况。

class Solution {    List<List<Integer>> result = new ArrayList<List<Integer>>();    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        Arrays.sort(candidates);        int length = candidates.length;        for(int i = 0 ; i < length ; i++){            //去除外围重复情况            if(i > 0 && candidates[i] == candidates[i-1]){continue;}            if(candidates[i] == target){                result.add(Arrays.asList(candidates[i]));            }else{                List<Integer> temp = new ArrayList<Integer>();                temp.add(candidates[i]);                combinationSum2(candidates, target-candidates[i], i + 1, temp);            }        }        return result;    }    public void combinationSum2(int[] candidates, int target, int startAt, List<Integer> currentList){        for(int i = startAt ; i<candidates.length ; i++){            if(candidates[i] == target){                currentList.add(candidates[i]);                result.add(currentList);                return;            }            if(candidates[i] > target){                return;            }            if(candidates[i] < target){                List<Integer> temp = new ArrayList<Integer>(currentList);                temp.add(candidates[i]);                combinationSum2(candidates, target-candidates[i], i + 1, temp);            }            //去除自遍历中的重复情况            while(i < candidates.length - 1 && candidates[i] == candidates[i + 1]){i++;}        }    }}