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);
}
}
}