39. Combination Sum

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();
        }
    }
}

原文 

https://segmentfault.com/a/1190000018118507

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