LeetCode 114 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]

分析：

看到返回所有，应该想到回溯，即DFS，DFS就要注意恢复现场。

public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(candidates==null || candidates.length==0)
return res;
//排序保证结果按递增排列
Arrays.sort(candidates);
dfs(candidates, 0, target, new ArrayList<Integer>(), res);
return res;
}

public void dfs(int[] candidates, int start, int target, List<Integer> item, List<List<Integer>> res){
//递归终结
if(target <= 0){
if(target==0)
res.add(new ArrayList<Integer>(item));
return;
}

for(int i=start; i<candidates.length; i++){
//因为i-1位置的元素已经用过多次，如果相等的话，i位置就不应该再处理
if(i>0 && candidates[i]==candidates[i-1])
continue;
item.add(candidates[i]);
//i没有增加，保证i位置的元素可以用多次，当target减小
dfs(candidates, i, target-candidates[i], item, res);
item.remove(item.size()-1);
}
}
}