Leetcode: 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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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] 

NP问题，遇到很多次了，这道题跟Combination Sum很像

第二遍方法：注意17行的跳过条件, 我在这里做错过，i > starter 才行， 仅仅i > 0是不行的。 i > starter限制的是一个数的取值不要重复，比如5,5,7,8，第一个数已经取过第一个5了，它就不要再尝试第二个5，但第二个数可以去尝试第二个5，所以第一个数取5同时第二个数也取5是可以的。但是如果把条件写成i > 0，第一个数取了第一个5，第二个数就连第二个5都取不了了，就没有第一个取5第二个也取5这种情况了

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> item = new ArrayList<Integer>();
        Arrays.sort(num);
        helper(res, item, num, target, 0);
        return res;
    }
    
    public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> item, int[] num, int remain, int starter) {
        if (remain < 0) return;
        if (remain == 0) {
            res.add(new ArrayList<Integer>(item));
            return;
        }
        for (int i=starter; i<num.length; i++) {
            if (i>starter && num[i] == num[i-1]) continue;
            item.add(num[i]);
            helper(res, item, num, remain-num[i], i+1);
            item.remove(item.size()-1);
        }
    }
}

Naive方法：

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (num == null || num.length == 0) return res;
        ArrayList<Integer> path = new ArrayList<Integer>();
        Arrays.sort(num);
        helper(res, path, num, target, 0, 0);
        return res;
    }
    
    public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path, int[] num, int target, int accum, int index) {
        if (accum > target) return;
        if (accum == target) {
            if (!res.contains(path)) res.add(new ArrayList<Integer>(path));
            return;
        }
        for (int i=index; i<num.length; i++) {
            path.add(num[i]);
            helper(res, path, num, target, accum+num[i], i+1);
            path.remove(path.size()-1);
        }
    }
}

CodeGanker的做法：注意21行他处理重复的情况，直接跳过

public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
    ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
    if(num == null || num.length==0)
        return res;
    Arrays.sort(num);
    helper(num,0,target,new ArrayList<Integer>(),res);
    return res;
}
private void helper(int[] num, int start, int target, ArrayList<Integer> item,
ArrayList<ArrayList<Integer>> res)
{
    if(target == 0)
    {
        res.add(new ArrayList<Integer>(item));
        return;
    }
    if(target<0 || start>=num.length)
        return;
    for(int i=start;i<num.length;i++)
    {
        if(i>start && num[i]==num[i-1]) continue;
        item.add(num[i]);
        helper(num,i+1,target-num[i],item,res);
        item.remove(item.size()-1);
    }
}