[LeetCode]040-Combination Sum II

最新推荐文章于 2021-03-19 10:57:05 发布

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该博客介绍了LeetCode第40题的解决方案，目标是找到一组候选数字的唯一组合，使得它们的和等于目标数字。在候选集C中，每个数字只能使用一次。例如，当候选数为[10,1,2,7,6,1,5]且目标数为8时，解包括[1,7]、[1,2,5]、[2,6]和[1,1,6]。解决思路涉及使用map进行除重操作。" 51126025,5633264,STM32f103 定时器配置与中断实战,"['嵌入式开发', 'STM32', '中断系统', 'C语言编程', '微控制器编程']

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

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

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

Solution:
思路，与上一题类似，不过在移位时，遍历时往后移一位，同时申请一个map用于除重。

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int> result;
        vector<vector<int>> results;
        map<vector<int>,int> checkmap;
        sort(candidates.begin(),candidates.end());
        combine(candidates,0,result,results,target,checkmap);

        return results;
    }

    void combine(vector<int>& candidates,int begin,vector<int>& result,vector<vector<int>>& results,int target,map<vector<int>,int>&checkmap)
    {
        int n =  candidates.size();
        int i = 0;
        for(i = begin;i<n;i++)
        {
            if(candidates[i] == target)
            {
                result.push_back(candidates[i]);
                if(checkmap.find(result) == checkmap.end())
                {
                    results.push_back(result);
                    checkmap.insert(pair<vector<int>,int>(result,i));   
                }
                result.pop_back();
                return;
            }
            else if(candidates[i] < target)
            {
                int  t = target - candidates[i];
                result.push_back(candidates[i]);
                combine(candidates,i+1,result,results,t,checkmap);
                if(!result.empty())
                {
                    result.pop_back();
                }
            }
            else
                return;
        }
    }
};