Leetcode -- Combination Sum

问题链接：https://oj.leetcode.com/problems/combination-sum/

问题描述：Given a set of candidate numbers (C) 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.

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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

问题API： public List<List<Integer>> combinationSum(int[] candidates, int target)

问题分析：

这题和Combination没啥不同，就是用一个buf保留每一层的数字，然后往下递归。只是每一位的数字都可以取无限次，所以可以在本地无限循环，然后再向下一层进行循环。另外，由于解集里的数字要顺序表示。所以最开始的数组就要sort一次。Arrays.sort()

    public List<List<Integer>> combinationSum(int[] candidates, int target) {        LinkedList<List<Integer>> res = new LinkedList<List<Integer>>();        Arrays.sort(candidates);        int size = target / candidates[0];        if(size == 0)            return res;        int[] tmpres = new int[size + 1];        combination(candidates, tmpres, target, 0, 0, res);        return res;    }        public void combination(int[] candidates, int[] tmpres, int target, int curpos, int curlevel, List<List<Integer>> res){        if(target == 0){            LinkedList<Integer> curres = new LinkedList<Integer>();            for(int i = 0; i < curpos; i++){                curres.add(tmpres[i]);            }            res.add(curres);        }else if(target < 0 || curlevel == candidates.length){            return;          }else{            for(int i = 0; target - candidates[curlevel] * i >= 0; i++){                if(i != 0){                    tmpres[curpos + i - 1] = candidates[curlevel];                }                combination(candidates, tmpres, target - i * candidates[curlevel], curpos + i, curlevel + 1, res);            }        }    }