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]
]

一刷
题解：
这里依然是用了跟上一题目很接近的方法。不同的地方在于，每个数字不可以被无限次。所以一个数只能一次，而且遇到重复数字我们要跳过。这样需要先sort数组，在for循环里要加入一条 - if (i > pos && candidates[i] == candidates[i - 1]) continue; 并且在DFS的时候每次

每次新的position = i + 1, 并不是上一题的position = I。

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

二刷
DFS + backtracking
for循环中，剔除的情况是，if(i!=index && candidates[i] == candidates[i-1]) continue;，因为在 candidates[i]的组合中，与candidates[i-1]的组合重合。（此时candidates[i-1]已经不在list中。

public class Solution {
    public List> combinationSum2(int[] candidates, int target) {
        List> res = new ArrayList<>();
        List list = new ArrayList<>();
        if(candidates == null || candidates.length == 0) return res;
        Arrays.sort(candidates);
        comb(res, list, 0, candidates, target);
        return res;
    }
    
    private void comb(List> res, List list, int index, int[] candidates, int target){
        if(target<0) return;
        if(target == 0){
            res.add(new ArrayList<>(list));
            return;
        }
        
        for(int i = index; i