40. Combination Sum II (JAVA)

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

与Combination Sum的区别在于，本题每次递归需要考虑重复元素，用while循环递归重复元素出现的次数；而Combination Sum每次递归只需要考虑两种情况，即放入该元素，或不放入该元素。

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<Integer> ans = new ArrayList<Integer>();
        Arrays.sort(candidates);
        backTrack(candidates, target, 0, ans, 0);
        return result;
    }

    public void backTrack(int[] candidates, int target, int start, List<Integer> ans, int sum){
        if(sum == target ){ //found an answer
            List<Integer> new_ans = new ArrayList<Integer>(ans); //不能用List<Integer> new_ans = ans;这个只是创建了原List的一个引用
            result.add(new_ans);
        }
        else if(start >= candidates.length || sum > target)
            return; //not found
        else{
            int cnt = 0; //repeated times
            while(start+1 < candidates.length && candidates[start+1]==candidates[start]){
                start++;
                cnt++;
            }
            // not choose current candidate
            backTrack(candidates,target,start+1,ans,sum);

            //choose current candidate
            List<Integer> backup = new ArrayList<Integer>(ans);
            int i = 0;
            for(i = 0; i <= cnt && sum <= target;i++){
                backup.add(candidates[start]);
                sum += candidates[start];
                backTrack(candidates,target,start+1,backup,sum);
            }
        }
    }

    private List<List<Integer>> result = new ArrayList<List<Integer>>();
}