Leetcode - Combination Sum II

My code:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {
    public List> combinationSum2(int[] candidates, int target) {
        ArrayList> result = new ArrayList>();
        if (candidates == null || candidates.length == 0)
            return result;
        Arrays.sort(candidates);
        ArrayList combinations = new ArrayList();
        if (candidates[0] > target)
            return result;
        else if (candidates[0] == target) {
            combinations.add(candidates[0]);
            result.add(combinations);
            return result;
        }
        boolean[] isVisited = new boolean[candidates.length];
        getCombinations(0, 0, candidates, target, isVisited, combinations, result);
        return result;
    }
        
    private void getCombinations(int begin, int sum, int[] candidates, int target, boolean[] isVisited,
                        ArrayList combinations, ArrayList> result) {
        for (int i = begin; i < candidates.length; i++) {
            int temp = candidates[i] + sum;
            if (temp > target)
                break;
            else if (temp == target) {
                combinations.add(candidates[i]);
                result.add(new ArrayList(combinations));
                combinations.remove(combinations.size() - 1);
                break;
            }
            else {
                isVisited[i] = true;
                if ( i >= 1 && candidates[i] == candidates[i - 1] && !isVisited[i - 1]) {
                    isVisited[i] = false;
                    continue;
                }
                combinations.add(candidates[i]);
                getCombinations(i + 1, temp, candidates, target, isVisited, combinations, result);
                combinations.remove(combinations.size() - 1);
                isVisited[i] = false;
            }
        }
            
    }
}

My test result:

Paste_Image.png

这道题目和之前的差不多，唯一的不同就是，只能用一次了，那么，就传递 i + 1就行了。
但是还需要考虑一个问题， candidates[] 数组中的元素是会有重复的，导致最后的结果也有重复，所以需要采用措施把这个重复情况排除掉，这个也是清楚怎么做的。

**
总结： Array, DFS
**

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public List> combinationSum2(int[] candidates, int target) {
        ArrayList> ret = new ArrayList>();
        if (candidates == null || candidates.length == 0 || target < 0)
            return ret;
        Arrays.sort(candidates);
        ArrayList group = new ArrayList();
        helper(0, 0, target, candidates, ret, group);
        return ret;
    }
    
    private void helper(int begin, int sum, int target, int[] candidates,
                            ArrayList> ret, ArrayList group) {
        if (sum > target)
            return;
        else if (sum == target) {
            ret.add(new ArrayList(group));
            return;
        }
        else {
            for (int i = begin; i < candidates.length; i++) {
                if (i > begin && candidates[i] == candidates[i - 1]) // in order to prevent repeating situation
                                                                    // such as, [1, 1] for 1
                    continue;
                group.add(candidates[i]);
                helper(i + 1, sum + candidates[i], target, candidates, ret, group);
                group.remove(group.size() - 1);
            }            
        }
    }
}

没能一遍过，一个地方错了。
就是这里是不允许重复的。
那么，如果[1, 1] for 1, 答案只能是一个[1].
所以需要加一个判断避免重复对相同情况DFS

Anyway, Good luck, Richardo!