回溯法总结

一般回溯法可以用两种框架，一种遍历方式(for循环)，选择方式（可以理解成到某一节点选择或者不选）。

比较二者的差别：

1.采用遍历方式，for（int i=dep;i选择的方式记得判断dep是否到达边界dep==nums.size()同时记得dep++;

2.遍历方式中for循环的临时变量存储的是temp[i],选择方式中是temp[dep];

3.采用选择方式中，pop完之后还需要继续进行回溯，而for循环中不需要，因为。

一个例子：

Combination Sum

Given a set of candidate numbers (C) 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.
• 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 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

方法1：

class Solution {
public:
	vector>res;
	vectortemp;
	void backtracking(int dep, int sum, vector& candidates)
	{
		if (sum==0)//子集和为0
		{
			res.push_back(temp);
			return;
		}
		if (sum < 0) return;
		for (int i = dep; i < candidates.size(); i++)
		{
			temp.push_back(candidates[i]);
			backtracking(i, sum - candidates[i], candidates);
			temp.pop_back();
			
		}
		

	}
	vector> combinationSum(vector& candidates, int target) {
		sort(candidates.begin(), candidates.end());
		backtracking(0, target, candidates);
		return res;
	}
};

方法二：

class Solution {
	public:
		vector>res;
		vectortemp;
		void backtracking(vector& candidates, int target,int dep)
		{
			if (target == 0)
			{
				res.push_back(temp);
				return;
			}
			if (target < 0) return;
			if (dep == candidates.size())return;//若不采用for循环的方式加上dep边界
			temp.push_back(candidates[dep]);
			backtracking(candidates, target - candidates[dep], dep); 
			temp.pop_back();
			target += candidates[dep];//归位
			backtracking(candidates, target - candidates[dep], dep+1);//若不采用for循环的方式记得pop完继续回溯
	
		}
		vector> combinationSum(vector& candidates, int target)
		{
			backtracking(candidates, target, 0);
			return res;
		}
};

Subsets

Given a set of distinct integers, nums, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

方法1：

class Solution {
public:
	vector>res;
	vectortemp;
	void backtracking(vector&nums, int pos)
	{
		res.push_back(temp);
		for (int i = pos; i < nums.size(); i++)
		{
			temp.push_back(nums[i]);
			backtracking(nums,i+1);
			temp.pop_back();
		}
	}
	vector> subsets(vector& nums)
	{
		sort(nums.begin(), nums.end());
		if (nums.size() == 0) return res;
		backtracking(nums, 0);
		return res;
	}
};

方法2：

class Solution {
	public:
		vector>res;
		vectortemp;
		void backtracking(vector&nums,int dep)
		{
			if (dep == nums.size())
			{
				res.push_back(temp);
				return;
			}
			temp.push_back(nums[dep]);
			backtracking(nums, dep + 1);
			temp.pop_back();
			backtracking(nums, dep + 1);
		}
		vector> subsets(vector& nums)
		{
		    sort(nums.begin(), nums.end());
			if (nums.size() == 0) return res;
			backtracking(nums, 0);
			return res;
		}
};