Combination Sum

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] 

思路：

简单的回溯法(递归实现).

比如对于数组3,2,6,7,target = 7,对数组排序得到[2,3,6,7]

1、第1个数字选取2, 那么接下来就是解决从数组[2,3,6,7]选择数字且target = 7-2 = 5

2、第2个数字选择2，那么接下来就是解决从数组[2,3,6,7]选择数字且target = 5-2 = 3

3、第3个数字选择2，那么接下来就是解决从数组[2,3,6,7]选择数字且target = 3-2 = 1

4、此时target = 1小于数组中的所有数字，失败，回溯，重新选择第3个数字

5、第3个数字选择3，那么接下来就是解决从数组[2,3,6,7]选择数字且target = 3-3 = 0

6、target = 0，找到了一组解，继续回溯寻找其他解

public List<List<Integer>> combinationSum(int[] candidates, int target) {ArrayList<List<Integer>> result = new ArrayList<List<Integer>>();if(candidates == null) {return result;}ArrayList<Integer> temp = new ArrayList<Integer>();Arrays.sort(candidates);combinationSumSolution(candidates, target, 0, result, temp);return result;    }private void combinationSumSolution(int[] candidates, int target, int index,ArrayList<List<Integer>> result, ArrayList<Integer> temp) {if(target == 0) {result.add(new ArrayList<Integer>(temp));return ;}for(int i=index; i<candidates.length &&target>=candidates[i] ; i++) {if(0==i || candidates[i]!= candidates[i-1]) {//忽略数组中的重复数字temp.add(candidates[i]);combinationSumSolution(candidates, target-candidates[i], i, result, temp);temp.remove(temp.size()-1);}}}