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Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.The same repeated number may be chosen from candidates unlimited number of times.Note:All numbers (including target) will be positive integers.The solution set must not contain duplicate combinations.Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
难度：medium
题目：给定一无重复元素的集合和一指定数，找出所有由数组内元素之和为该数的序列。每个元素都可以被无限次使用。注意：所以元素都为正整数包括给定的整数。答案不允许有重复的组合。
思路：递归
Runtime: 9 ms, faster than 81.90% of Java online submissions for Combination Sum.Memory Usage: 29.6 MB, less than 16.71% of Java online submissions for Combination Sum.
class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> result = new ArrayList();
combinationSum(candidates, 0, target, 0, new Stack<Integer>(), result);
return result;
}
public void combinationSum(int[] cs, int idx, int target, int sum, Stack<Integer> stack, List<List<Integer>> result) {
if (sum == target) {
result.add(new ArrayList(stack));
return;
}
for (int i = idx; i < cs.length; i++) {
if (sum + cs[i] <= target) {
stack.push(cs[i]);
combinationSum(cs, i, target, sum + cs[i], stack, result);
stack.pop();
}
if (sum + cs[i] >= target) break;
}
}
}
