46 . 全排列
链接 :
. - 力扣（LeetCode）
思路 :
那么怎么确定选了那个数呢?
这里设置一个used表示i选没选过 ;
class Solution {
public:vector<vector<int>> ans;vector<int> path;void backtrack(vector<int>nums,vector<bool> used){if(path.size() >= nums.size()){ans.push_back(path);return;}for(int i=0;i<nums.size();i++){if(used[i]==true) continue;used[i] = true;path.push_back(nums[i]);backtrack(nums,used);used[i] = false;path.pop_back();}}vector<vector<int>> permute(vector<int>& nums) {ans.clear();path.clear();vector<bool> used(nums.size(),false);backtrack(nums,used);return ans;}
};51 . N皇后
链接 :
. - 力扣（LeetCode）
思路 :
class Solution {
public:vector<vector<string>> solveNQueens(int n) {vector<vector<string>> ans;vector<int> col(n), on_path(n), diag1(n * 2 - 1), diag2(n * 2 - 1);function<void(int)> dfs = [&](int r) {if (r == n) {vector<string> board(n);for (int i = 0; i < n; ++i)board[i] = string(col[i], '.') + 'Q' + string(n - 1 - col[i], '.');ans.emplace_back(board);return;}for (int c = 0; c < n; ++c) {int rc = r - c + n - 1;if (!on_path[c] && !diag1[r + c] && !diag2[rc]) {col[r] = c;on_path[c] = diag1[r + c] = diag2[rc] = true;dfs(r + 1);on_path[c] = diag1[r + c] = diag2[rc] = false; // 恢复现场}}};dfs(0);return ans;}
};
52. N 皇后 II
链接 :
. - 力扣（LeetCode）
思路 :
直接套用上一题代码,返回ans.size()即可 ;
代码 :
class Solution {
public:int totalNQueens(int n) {vector<vector<string>> ans;vector<int> col(n), on_path(n), diag1(n * 2 - 1), diag2(n * 2 - 1);function<void(int)> dfs = [&](int r) {if (r == n) {vector<string> board(n);for (int i = 0; i < n; ++i)board[i] = string(col[i], '.') + 'Q' + string(n - 1 - col[i], '.');ans.emplace_back(board);return;}for (int c = 0; c < n; ++c) {int rc = r - c + n - 1;if (!on_path[c] && !diag1[r + c] && !diag2[rc]) {col[r] = c;on_path[c] = diag1[r + c] = diag2[rc] = true;dfs(r + 1);on_path[c] = diag1[r + c] = diag2[rc] = false; // 恢复现场}}};dfs(0);return ans.size();}
};视频链接 :
回溯算法套路③排列型回溯+N皇后【基础算法精讲 16】_哔哩哔哩_bilibili
