Combination Sum I II之回溯解法

题目

题目I

Given a set of candidate numbers (C) (without duplicates) and a target number (T),
find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.

题目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.

分析

题目I是在一个没有重复数字，但是数字可以重复选择的数组中找到所有累加和为target的集合，集合中不考虑位置且结果不能有重复。
题目II是在一个有重复数字，但是每个数字只能选择一次的数组中找到所有累加和为target的集合，集合中不考虑位置且结果不能有重复。
还有一种动态规划解法，这里先用回溯。

代码

I

public List> combinationSum(int[] candidates, int target) {
    List> res = new ArrayList();
    combinationSum(candidates, target, new ArrayList(), res, 0);
    return res;
}

private void combinationSum(int[] candidates, int target, List list, List> res, int start){
    if(target > 0){
        for(int i = start; i < candidates.length; ++i){
            list.add(candidates[i]);
            //回溯还是从i开始，因为可以重复选取。
            combinationSum(candidates, target - candidates[i],  list, res, i);
            list.remove(list.size() - 1);
        }
    }else if(target == 0){
        res.add(new ArrayList(list));
    }
}

II

public List> combinationSum2(int[] candidates, int target) {
    Arrays.sort(candidates);    //先排序
    List> res = new ArrayList();
    if(candidates == null || candidates.length == 0 || target <= 0) return res;
    helper(candidates, res, new ArrayList(), target, 0);
    return res;
}

public void helper(int[] candidates, List> res, List list, int target, int start){
    if(target == 0){
        res.add(new ArrayList(list));
    }else if(target > 0){
        for(int i = start; i < candidates.length; ++i){
            if(i > start && candidates[i] == candidates[i-1]){  //防止重复，I不存在这个问题
                continue; 
            }
            
            list.add(candidates[i]);
            helper(candidates, res, list, target - candidates[i], i + 1);//从下一个数开始
            list.remove(list.size() - 1);
        }
    }
}