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题目描述：
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:
For example, given candidate set 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
解题思路：
首先将数组排序，然后应用回溯和贪心的算法，首先从最小的数开始，尽可能的将该数字放入List中，如果加入的数目太多导致下面没有数字可以使综合等于target则将之前加入的数字弹出，换进下一个数字，如此反复。
代码如下：
public class Solution {
public List<List<Integer>> combinationSum(int[] candidates,
int target) {
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList<List<Integer>>();
getResult(result, new ArrayList<Integer>(), candidates, target, 0);
return result;
}
public void getResult(List<List<Integer>> result,
List<Integer> current, int[] candiates, int target, int start) {
if (target > 0) {
for (int i = start; i < candiates.length && target >= candiates[i]; i++) {
current.add(candiates[i]);
getResult(result, current, candiates, target - candiates[i], i);
current.remove(current.size() - 1);
}
} else if (target == 0) {
result.add(new ArrayList<Integer>(current));
}
}
}
Java [Leetcode 39]Combination Sum
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原文地址：http://www.cnblogs.com/zihaowang/p/5020537.html
