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leetcode(40) combination sum ii 解法

admin 11月 29, 2020 0

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

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8
 
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

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Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

解法

这题只需要在Leetcode(39)上做一点点的改动，这里不允许一个数重复利用，故只需要将下一次搜索的起点改为i+1即可，同时考虑去除重复输出的情况，即在start之后进行分叉搜索时，若nums[i] = nums[i-1]时，说明已经考虑过了，直接进入下一次循环。

具体代码如下：

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class  {
    List<List<Integer>> res = new ArrayList<>();

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<Integer> tmp = new ArrayList<>();
        Arrays.sort(candidates);
        helper(candidates, target, tmp, 0);
        return res;
    }

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

    }
}
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