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。一个元素可以被无限次使用。
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: 15 ms, faster than 33.80% of Java online submissions for Combination Sum.
Memory Usage: 25.8 MB, less than 64.72% of Java online submissions for Combination Sum.
class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> result = new ArrayList<>(); combinationSum(candidates, 0, 0, target, new Stack<Integer>(), result); return result; } private void combinationSum(int[] cs, int curIdx, int sum, int target, Stack<Integer> stack, List<List<Integer>> result) { if (curIdx >= cs.length || sum > target) { return; } if (sum == target) { result.add(new ArrayList<>(stack)); return; } for (int i = curIdx; i < cs.length; i++) { stack.push(cs[i]); combinationSum(cs, i, sum + cs[i], target, stack, result); stack.pop(); } } }