Combination Sum 135
Question
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.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
Notice
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.
Solution
与Subsets类似。注意几点:
1) 每加入一个数,就要把下一层的targte为原来的targte减去这个数
2) 需要考虑避免重复:对元素排序,判断是否和前一个数等值
3) 可以重复取相同元素,所以下一层的起始pos和上一层相同
代码如下:
public class Solution {
/**
* @param candidates: A list of integers
* @param target:An integer
* @return: A list of lists of integers
*/
public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
// write your code here
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if(candidates == null || candidates.length == 0 || target <= 0){
return result;
}
ArrayList<Integer> list = new ArrayList<Integer>();
Arrays.sort(candidates);
helper(result, list, candidates, target, 0);
return result;
}
private void helper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, int[] candidates, int target, int pos){
if(target == 0){
result.add(new ArrayList<Integer>(list));
return;
}
for(int i = pos; i < candidates.length; i++){
if(i != 0 && candidates[i] == candidates[i - 1]){
continue;
}
if(candidates[i] > target){
return;
}
list.add(candidates[i]);
helper(result, list, candidates, target - candidates[i], i);
list.remove(list.size() - 1);
}
}
}