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  • 【LeetCode】Combination Sum II

    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 (a1a2, � , 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] 

    思路:

    同样DFS,跟上一题是一样的,只不过为了不重复,每次调用dfs的时候pos = i + 1;

    这道题备选数组是有重复的,得去重

    代码:

    class Solution {
    public:
        vector<vector<int> > combinationSum2(vector<int> &num, int target) {
            vector<int> temp;
            vector<vector<int> > rst;
            sort(num.begin(), num.end());
            dfs(num, target, temp, rst, 0);
            return rst;
        }
    
        void dfs(vector<int> &num, int target, vector<int> &temp, vector<vector<int> > &rst, int pos)
        {
            if (target < 0)
                return;
            if (target == 0)
            {
                rst.push_back(temp);
                return;
            }
    
            for (int i = pos; i < num.size(); i++)
            {
                temp.push_back(num[i]);
                dfs(num, target - num[i], temp, rst, i + 1);
                temp.pop_back();
                while (i < num.size() - 1 && num[i] == num[i+1])
                    i++;
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/dollarzhaole/p/3243254.html
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