40. Combination Sum II

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]
]

 

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        if(candidates.length==0) return new ArrayList();
        Arrays.sort(candidates);
        List<List<Integer>> results = new ArrayList();
        List<Integer> result = new ArrayList();
        dfs(candidates, results, result, target, 0);
        return results;
    }
    public void dfs(int[] candidates,List<List<Integer>> results,List<Integer> result, int target, int start ){
        if(target==0){
            System.out.println(result);
            results.add(new ArrayList(result));
            return;
        }
        int pre = -1;

        for(int i = start; i < candidates.length; i++){
            if(pre==candidates[i]) continue;//用previous防止重复
            if((target-candidates[i]) < 0){
                return;
            }
            pre = candidates[i];
            result.add(candidates[i]);
            dfs(candidates, results, result, target-candidates[i],i+1);//i变成i+1
            result.remove(result.size()-1);
        }
    }
}

     // 如果上一轮循环已经使用了nums[i]，则本次循环就不能再选nums[i]，
            // 确保nums[i]最多只用一次

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
            List<List<Integer>> list = new ArrayList<>();
    backtrack(list, new ArrayList<>(), candidates, target, 0);
    return list;
    }
    private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) list.add(new ArrayList<>(tempList));
    else{ 
        for(int i = start; i < nums.length; i++){
            if((i > start) && nums[i] == nums[i-1]) continue;
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, remain - nums[i], i + 1); 
            //找到了一个解或者 remain < 0 了，将当前数字移除，然后继续尝试
            tempList.remove(tempList.size() - 1);
        }
    }
}
}

首先在递归的for循环里加上if (i > start && num[i] == num[i - 1]) continue; 这样可以防止res中出现重复项，然后就在递归调用combinationSum2DFS里面的参数换成i+1，这样就不会重复使用数组中的数字了

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList();
        Arrays.sort(candidates);
        help(res, new ArrayList(), 0, candidates, target, 0);
        return res;
    }
    
    public void help(List<List<Integer>> res, List<Integer> cur, int count, int[] candidates, int target, int start) {
        if(count > target) return;
        if(count == target) {
            res.add(new ArrayList(cur));
            return;
        }
        
        for(int i = start; i < candidates.length; i++) {
            if(i > start && candidates[i] == candidates[i - 1]) continue;
            count += candidates[i];
            cur.add(candidates[i]);
            help(res, cur, count, candidates, target, i + 1);
            count -= candidates[i];
            cur.remove(cur.size() - 1);
        }
    }
}

感觉最重要的就是如何判断不加入重复，这里我们设定，当在同一个循环中，i 》start且i与前面的数字重复后，我们跳过。但如果i==start时就不用管重复，因为刚开始了一个新循环。

1，1，2，5.不判断会有两个1，2，5，判断后只有一个1，2，5。