【LeetCode】46. Permutations（C++）

地址：https://leetcode.com/problems/permutations/

题目：

Given a collection of distinct integers, return all possible permutations.
**Example **:

理解：

和combination number差不多的思路，用backtracking。直接看代码吧，差别不大。

实现1：

来源：Share My C++ backtracksolution

class Solution {
public:
	vector<vector<int>> permute(vector<int>& nums) {
		vector<vector<int>> res;
		vector<int> curr;
		vector<bool> flags(nums.size(), false);
		backtracking(nums, res, curr, flags);
		return res;
	}

	void backtracking(vector<int>& nums, vector<vector<int>>& res, vector<int>& curr, vector<bool>& flags) {
		if (nums.size() == curr.size()) {
			res.push_back(curr);
		}
		else {
			for (int i = 0; i < nums.size(); ++i) {
				if (flags[i])
					continue;
				curr.push_back(nums[i]);
				flags[i] = true;
				backtracking(nums, res, curr,flags);
				flags[i] = false;
				curr.pop_back();
			}
		}
	}
};

实现2：

来源：My elegant recursive C++ solution with inline explanation

这里有比较直观的解释：https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

class Solution {
public:
	vector<vector<int>> permute(vector<int>& nums) {
		vector<vector<int>> res;
		backtracking(nums, res, 0);
		return res;
	}

	void backtracking(vector<int>& nums, vector<vector<int>>& res,int begin) {
		if (begin>=nums.size()) {
			res.push_back(nums);
		}
		else {
			for (int i = begin; i < nums.size(); ++i) {
				swap(nums[begin], nums[i]);
				backtracking(nums, res, begin + 1);
				swap(nums[begin], nums[i]);
			}
		}
	}
};

果然人与人之间的差距比人与狗都大。。