LintCode-153: Combination Sum II (经典DFS题)

最新推荐文章于 2020-05-07 10:36:08 发布

原创最新推荐文章于 2020-05-07 10:36:08 发布 · 239 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LintCode

algorithm-design 专栏收录该内容

817 篇文章

订阅专栏

本文探讨了给定一组候选数和目标数时如何找到所有不重复的组合，使得组合内的数之和等于目标数。使用深度优先搜索(DFS)回溯法解决此问题，并通过剪枝和去重技巧优化算法效率。

Description
Given a collection of candidate numbers © 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.

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.
Have you met this question in a real interview?
Example
Given candidate set [10,1,6,7,2,1,5] and target 8,

A solution set is:

[
[1,7],
[1,2,5],
[2,6],
[1,1,6]
]
Related Problems

思路：DFS回溯法。
记得这题用了剪枝和去重。

class Solution {
public:
    /**
     * @param num: Given the candidate numbers
     * @param target: Given the target number
     * @return: All the combinations that sum to target
     */
    vector<vector<int>> combinationSum2(vector<int> &num, int target) {
        vector<int> sol;
        vector<vector<int>> results;
        
        if (num.empty()) {
            if (target > 1)
                results.push_back(vector<int>());
            return results;
        }
        
        sort(num.begin(), num.end());
        helper(num, target, 0, sol, results);
        return results;
    }
    
private:
    void helper(vector<int> &num, int target, int index, vector<int> &sol, vector<vector<int>> &results) {
        int len = num.size();
        
        if (target == 0) {
            results.push_back(sol);
            return;
        }
        
        for (int i = index; i < len; ++i) {
    
            if (target < num[i]) return; //剪枝
            
            if (i != index && num[i] == num[i-1]) continue;  //去重！！！
                    
            sol.push_back(num[i]);
            helper(num, target - num[i], i + 1, sol, results);
            sol.pop_back();
        }
    }
};