leetcode || 40、Combination Sum II

最新推荐文章于 2025-08-16 00:47:01 发布

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本文详细介绍了如何使用深度优先搜索（DFS）算法解决组合总和 II 问题，即从给定候选数集中找到所有唯一组合，这些组合的元素之和等于给定的目标值，且每个数只能使用一次。通过排序、递归和回溯，有效地解决了这个问题，并提供了代码实现。

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problem:

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]

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Array Backtracking

是上一题的变形，这里每个数字只允许使用一次

thinking :

(1)DFS算法，路径深度每次+1，但是否采用该层数字则不一定！！！

code:

class Solution {
private:
    vector<vector<int> > ret;
    vector<int> a;
public:
    void solve(int dep, int maxDep, vector<int> &num, int target)
    {
        if (target < 0)
            return;
            
        if (dep == maxDep)
        {
            if (target == 0)
            {
                vector<int> res;
                for(int i = 0; i < maxDep; i++)
                    for(int j = 0; j < a[i]; j++)
                        res.push_back(num[i]);
                ret.push_back(res);
            }
            
            return;
        }
        
        for(int i = 0; i <= min(target / num[dep], 1); i++)
        {
            a[dep] = i;
            
            if (i == 1 && dep > 0 && num[dep-1] == num[dep] && a[dep-1] == 0)
                continue;
                
            solve(dep + 1, maxDep, num, target - i * num[dep]);
        }
    }
    
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(num.begin(), num.end());
        a.resize(num.size());
        ret.clear();
        if (num.size() == 0)
            return ret;
            
        solve(0, num.size(), num, target);
        
        return ret;
    }
};