[LeetCode] 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]
]

组合总和。这个题依然是backtracking系列里面需要背下来的题目。既然是求什么样的组合的sum能等于target，那么helper函数的退出条件就是看看什么样的组合的sum == target。同时，为了剪枝/加速，如果当前的sum大于target，就可以提前终止回溯了。

注意21行为什么递归到下一层的时候还是从i开始是因为数字可以被重复利用。这个地方跟40题还是有点区别的。

时间O(2^n)

空间O(n)

Java实现

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if (candidates == null || candidates.length == 0) {
            return res;
        }
        helper(res, new ArrayList<>(), candidates, target, 0);
        return res;
    }

    private void helper(List<List<Integer>> res, List<Integer> list, int[] candidates, int target, int start) {
        if (target < 0) {
            return;
        }
        if (target == 0) {
            res.add(new ArrayList<>(list));
            return;
        }
        for (int i = start; i < candidates.length; i++) {
            list.add(candidates[i]);
            helper(res, list, candidates, target - candidates[i], i);
            list.remove(list.size() - 1);
        }
    }
}

LeetCode 题目总结