LeetCode的39和40两道题都是CombinationSum,它们都可以用深度优先搜索(DFS)来解决(当然也还有许多其它方法)。这篇博客就系统的来介绍DFS和CombinationSum这一类问题。

题目

LeetCode39:

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[ [7], [2, 2, 3] ]

LeetCode40:

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.
  • 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] ]

分析

39和40的区别在于,对于集合中的每个元素,39中每个元素可以取多次,40中每个元素只可取一次(但是元素的数值可重复)

深度优先搜索

形象的用下面的图像表示一个序列[a b c]在搜索时的搜索路径,对于第一个问题:

对于第二个问题:

它们的剪枝条件都是: sum(selected) >= target

其中第二个问题,由于数字可能重复,所以需要先进行排序,然后通过一些方法去除排列组合引起的重复,具体方法在代码分析中会讲到。

代码

CombinationSum

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:

	vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
		if (candidates.empty()) return res;
		dfs(0, 0, target, candidates);
		return res;
	}
private
	void dfs(int start, int sum, int target, vector<int>& candidates){
		if (sum == target){
			res.push_back(ans);
			return;
		} else if (sum > target){
			return;
		} else{
			for (int i = start; i < candidates.size(); i++){
				ans.push_back(candidates[i]);
				dfs(i, sum + candidates[i], target, candidates);
				ans.pop_back();
			}
		}

	}
	vector<vector<int>>res;
	vector<int>ans;
};

int main(int argc, char *argv[]){
	Solution s;
	system("pause");
	return 0;
}

CombinationSum2

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

代码分析

第二份代码首先进行了排序,一方面是出于效率考虑,另一方面也是判断重复数字的需要。

由于每个数字不能重复,在每层cur时,第一份代码是由i开始的而第二份代码是从i+1开始的。同样是由于不能重复,所以有下面一行代码:

if (cur != i&&candidates[i] == candidates[i-1]) continue;

我们举一个例子来说明下:

如果一个序列是 1 1 1 1 6, target = 8

那么在扫面第一个1时,将后面的第2,3,4号,三个1都看做一个1,因为后面的1选择哪一个都是等价的,如果不加区分会出现重复,按照这样的规则进行搜索,就可以避免重复。这是第二个问题中最核心的一段代码。

 

 

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