39. Combination Sum

深碍√TFBOYSˉ_ 2022-03-07 03:35 170阅读 0赞

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;
            }
        }
    }