【Leetcode】Combination Sum

题目链接：https://leetcode.com/problems/combination-sum/

题目：

C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

same repeated number may be chosen from C

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.

2,3,6,7 and target 7, 

A solution set is: 

[7] 

[2, 2, 3]

思路：

允许重复，但选择的数的序列必须是有序的。

算法：

int cl[] = null;
	List<List<Integer>> iLists = new ArrayList<List<Integer>>();
	List<Integer> list = new ArrayList<Integer>();
	int target = 0;

	public List<List<Integer>> combinationSum(int[] candidates, int target) {
		Arrays.sort(candidates);
		this.cl = candidates;
		this.target = target;
		dspCombination(0, 0);
		return iLists;
	}

	private void dspCombination(int sum, int level) {
		if (sum == target) {
			iLists.add(new ArrayList<Integer>(list));
			return;
		} else if (sum > target) {
			return;
		} else {
			//类似于迷宫问题
			for (int i = level; i < cl.length; i++) { //i=level保证了下一步不会选择[level]之前的数，即保证了是有序的
				sum += cl[i];
				list.add(cl[i]);
				dspCombination(sum, i); //没有用dspCombination(sum,i+1)给了当前cl[i]再次驻留机会，即允许重复
				list.remove(list.size() - 1);
				sum -= cl[i];
			}
		}
	}