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  • LeetCode 40. Combination Sum II 组合总和 II (C++/Java)

    题目:

    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:

    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:

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

    分析:

    本题是LeetCode 39. Combination Sum 组合总和 (C++/Java)的扩展,我们还是利用39题中的方法来做这道题。

    只不过本题所给的数字有重复,且要求最后的结果中不能有重复的,例如例子1,有两个1,他们都可以和7组合成8,如果按照39题搜索的方法就会产生重复结果。

    其中一个办法就是将搜索生成的结果加入到set中,这样集合类会自动帮我们去重。

    此外我们还可以在每一轮搜索时,当发现有相同元素时,直接跳过本次搜素,这样就不会产生重复的结果了。

    程序:

    C++

    class Solution {
    public:
        vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
            vector<vector<int>> res;
            vector<int> curr;
            sort(candidates.begin(), candidates.end());
            dfs(candidates, target, 0, res, curr);
            return res;
        }
        void dfs(vector<int>& candidates, int target, int index, vector<vector<int>>& res, vector<int> curr){
            if(target == 0){
                res.push_back(curr);
                return;
            }
            for(int i = index; i < candidates.size(); ++i){
                if(candidates[i] > target)
                    return;
                if(i > index && candidates[i] == candidates[i-1])
                    continue;
                curr.push_back(candidates[i]);
                dfs(candidates, target-candidates[i], i+1, res, curr);
                curr.pop_back();
            }
        }
    };

    Java

    class Solution {
        public List<List<Integer>> combinationSum2(int[] candidates, int target) {
            List<List<Integer>> res = new ArrayList<>();
            LinkedList<Integer> curr = new LinkedList<>();
            Arrays.sort(candidates);
            dfs(candidates, target, 0, res, curr);
            return res;
        }
        private void dfs(int[] candidates, int target, int index, List<List<Integer>> res, LinkedList<Integer> curr){
            if(target == 0){
                res.add(new LinkedList<>(curr));
                return;
            }
            for(int i = index; i < candidates.length; ++i){
                if(candidates[i] > target)
                    return;
                if(i > index && candidates[i] == candidates[i-1])
                    continue;
                curr.addLast(candidates[i]);
                dfs(candidates, target-candidates[i], i+1, res, curr);
                curr.removeLast();
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/silentteller/p/12372194.html
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