【算法篇】回溯算法类（2）（笔记）
目录
一、LeetCode 题目
1. 子集II
给你一个整数数组
nums，其中可能包含重复元素，请你返回该数组所有可能的子集（幂集）。解集 不能 包含重复的子集。返回的解集中，子集可以按 任意顺序 排列。
示例 1：
输入：nums = [1,2,2]
输出：[[],[1],[1,2],[1,2,2],[2],[2,2]]
示例 2：
输入：nums = [0]
输出：[[],[0]]
// 方法一：
class Solution {
public:
vector<vector<int>> result;
vector<int> path;
void backtracking(vector<int>& nums, int startIndex) {
result.push_back(path);
for (int i = startIndex; i < nums.size(); i++) {
if (i > startIndex && nums[i] == nums[i - 1]) {
continue;
}
path.push_back(nums[i]);
backtracking(nums, i + 1);
path.pop_back();
}
return;
}
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
backtracking(nums, 0);
return result;
}
};
// 方法二：（unordered_set 的使用）
class Solution {
public:
vector<vector<int>> result;
vector<int> path;
void backtracking(vector<int>& nums, int startIndex) {
result.push_back(path);
unordered_set<int> uset;
for (int i = startIndex; i < nums.size(); i++) {
if (uset.find(nums[i]) != uset.end()) {
continue;
}
uset.insert(nums[i]);
path.push_back(nums[i]);
backtracking(nums, i + 1);
path.pop_back();
}
return;
}
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
backtracking(nums, 0);
return result;
}
};2. 递增子序列
给你一个整数数组
nums，找出并返回所有该数组中不同的递增子序列，递增子序列中 至少有两个元素 。你可以按 任意顺序 返回答案。数组中可能含有重复元素，如出现两个整数相等，也可以视作递增序列的一种特殊情况。
示例 1：
输入：nums = [4,6,7,7]
输出：[[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
示例 2：
输入：nums = [4,4,3,2,1]
输出：[[4,4]]
class Solution {
private:
vector<vector<int>> result;
vector<int> path;
void backtracking(vector<int>& nums, int startIndex) {
if (path.size() > 1) {
result.push_back(path);
}
unordered_set<int> uset; // 使用set对本层元素进行去重
for (int i = startIndex; i < nums.size(); i++) {
if ((!path.empty() && nums[i] < path.back())
|| uset.find(nums[i]) != uset.end()) {
continue;
}
uset.insert(nums[i]); // 记录这个元素在本层用过了，本层后面不能再用了
path.push_back(nums[i]);
backtracking(nums, i + 1);
path.pop_back();
}
}
public:
vector<vector<int>> findSubsequences(vector<int>& nums) {
backtracking(nums, 0);
return result;
}
};3. 全排列
给定一个不含重复数字的数组
nums，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
示例 1：
输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
示例 2：
输入：nums = [0,1]
输出：[[0,1],[1,0]]
示例 3：
输入：nums = [1]
输出：[[1]]
class Solution {
public:
vector<vector<int>> result;
vector<int> path;
void backtracking(vector<int>& nums, vector<bool>& used) {
if (path.size() == nums.size()) {
result.push_back(path);
return;
}
for (int i = 0; i < nums.size(); i++) {
if (used[i] == true) {
continue;
}
path.push_back(nums[i]);
used[i] = true;
backtracking(nums, used);
path.pop_back();
used[i] = false;
}
return;
}
vector<vector<int>> permute(vector<int>& nums) {
vector<bool> used(nums.size(), false);
backtracking(nums, used);
return result;
}
};4. 全排列 II
给定一个可包含重复数字的序列
nums，按任意顺序 返回所有不重复的全排列。
示例 1：
输入：nums = [1,1,2]
输出：
[[1,1,2],
[1,2,1],
[2,1,1]]
示例 2：
输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]class Solution {
public:
vector<vector<int>> result;
vector<int> path;
void backtracking(vector<int>& nums, vector<bool>& used) {
if (path.size() == nums.size()) {
result.push_back(path);
return;
}
for (int i = 0; i < nums.size(); i++) {
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
// 如果上一个数还没用，那么选择第i个和上一个没什么区别
continue;
}
if (used[i] == false) {
path.push_back(nums[i]);
used[i] = true;
backtracking(nums, used);
path.pop_back();
used[i] = false;
}
}
return;
}
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<bool> used(nums.size(), false);
sort(nums.begin(), nums.end());
backtracking(nums, used);
return result;
}
};树层上去重(used[ i - 1 ] == false)，的树形结构如下：
树枝上去重（used[ i - 1 ] == true）的树型结构如下：
5. 重新安排行程
给你一份航线列表
tickets，其中tickets[i] = [fromi, toi]表示飞机出发和降落的机场地点。请你对该行程进行重新规划排序。所有这些机票都属于一个从
JFK（肯尼迪国际机场）出发的先生，所以该行程必须从JFK开始。如果存在多种有效的行程，请你按字典排序返回最小的行程组合。
- 例如，行程
["JFK", "LGA"]与["JFK", "LGB"]相比就更小，排序更靠前。假定所有机票至少存在一种合理的行程。且所有的机票 必须都用一次 且 只能用一次。
示例 1：
输入：tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
输出：["JFK","MUC","LHR","SFO","SJC"]
示例 2：
输入：tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
输出：["JFK","ATL","JFK","SFO","ATL","SFO"]
解释：另一种有效的行程是 ["JFK","SFO","ATL","JFK","ATL","SFO"] ，但是它字典排序更大更靠后。class Solution {
private:
// unordered_map<出发机场, map<到达机场, 航班次数>> targets
unordered_map<string, map<string, int>> targets;
bool backtracking(int ticketNum, vector<string>& result) {
if (result.size() == ticketNum + 1) {
return true;
}
for (pair<const string, int>& target : targets[result[result.size() - 1]]) {
if (target.second > 0 ) { // 记录到达机场是否飞过了
result.push_back(target.first);
target.second--;
if (backtracking(ticketNum, result)) return true;
result.pop_back();
target.second++;
}
}
return false;
}
public:
vector<string> findItinerary(vector<vector<string>>& tickets) {
targets.clear();
vector<string> result;
for (const vector<string>& vec : tickets) {
targets[vec[0]][vec[1]]++; // 记录映射关系
}
result.push_back("JFK"); // 起始机场
backtracking(tickets.size(), result);
return result;
}
};6. N皇后
https://leetcode.cn/problems/n-queens/description/
https://leetcode.cn/problems/n-queens/description/
按照国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。n 皇后问题 研究的是如何将
n个皇后放置在n×n的棋盘上，并且使皇后彼此之间不能相互攻击。给你一个整数
n，返回所有不同的 n 皇后问题 的解决方案。每一种解法包含一个不同的 n 皇后问题 的棋子放置方案，该方案中'Q'和'.'分别代表了皇后和空位。
示例 1：
输入：n = 4
输出：[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
解释：如上图所示，4 皇后问题存在两个不同的解法。
示例 2：
输入：n = 1
输出：[["Q"]]
class Solution {
public:
vector<vector<string>> result;
void backtracking(int n, int row, vector<string>& chessboard) {
if (row == n) {
result.push_back(chessboard);
return;
}
for (int col = 0; col < n; col++) { // 每一行的不同位置
if (isValid(row, col, chessboard, n)) {
chessboard[row][col] = 'Q';
backtracking(n, row + 1, chessboard);
chessboard[row][col] = '.';
}
}
return;
}
bool isValid(int row, int col, vector<string>& chessboard, int n) {
// 比较列
for (int i = 0; i < row; i++) {
if (chessboard[i][col] == 'Q') return false;
}
// 比较45度
for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
if (chessboard[i][j] == 'Q') return false;
}
// 比较135度
for (int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
if (chessboard[i][j] == 'Q') return false;
}
return true;
}
vector<vector<string>> solveNQueens(int n) {
vector<string> chessboard(n, string(n, '.'));
backtracking(n, 0, chessboard);
return result;
}
};7. 解数独
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
- 数字
1-9在每一行只能出现一次。- 数字
1-9在每一列只能出现一次。- 数字
1-9在每一个以粗实线分隔的3x3宫内只能出现一次。（请参考示例图）数独部分空格内已填入了数字，空白格用
'.'表示。
示例 1：
输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
思路：
一个 for 循环遍历棋盘的行，一个for循环遍历棋盘的列，一行一列确定下来之后，递归遍历这个位置放9个数字的可能性
// 写法一：
class Solution {
public:
bool backtracking(vector<vector<char>>& board) {
for (int i = 0; i < board.size(); i++) { // 遍历行
for (int j = 0; j < board[0].size(); j++) { // 遍历列
if (board[i][j] == '.') {
for (char k = '1'; k <= '9'; k++) {
if (isValid(board, i, j, k)) {
board[i][j] = k;
if (backtracking(board)) return true; // 如果找到合适一组立刻返回
board[i][j] = '.';
}
}
return false; // 9个数都试完了，都不行，返回false
}
}
}
return true; // 遍历完没有返回false，说明找到了合适棋盘位置了
}
bool isValid(vector<vector<char>>& board, int row, int col, char key) {
// 检查行
for (int i = 0; i < 9; i++) {
if (board[row][i] == key) return false;
}
// 检查列
for (int i = 0; i < 9; i++) {
if (board[i][col] == key) return false;
}
// 检查四方格
int rows = row / 3;
int cols = col / 3;
for (int i = 3 * rows; i < 3 * rows + 3; i++) {
for (int j = 3 * cols; j < 3 * cols + 3; j++) {
if (board[i][j] == key) return false;
}
}
return true;
}
void solveSudoku(vector<vector<char>>& board) {
backtracking(board);
return;
}
};
// 写法二：
class Solution {
public:
bool backtracking(vector<vector<char>>& board) {
for (int i = 0; i < board.size(); i++) { // 遍历行
for (int j = 0; j < board[0].size(); j++) { // 遍历列
if (board[i][j] == '.') { // 递归的上一层已经填上一个数字了，所以这里这里直接跳过了
for (char k = '1'; k <= '9'; k++) {
if (isValid(board, i, j, k)) {
board[i][j] = k;
if (backtracking(board)) return true; // 如果找到合适一组立刻返回
board[i][j] = '.';
}
}
return false; // 9个数都试完了，都不行，返回false
}
}
}
return true; // 遍历完没有返回false，说明找到了合适棋盘位置了
}
bool isValid(vector<vector<char>>& board, int row, int col, char key) {
// 检查行
for (int i = 0; i < 9; i++) {
if (board[row][i] == key) return false;
}
// 检查列
for (int i = 0; i < 9; i++) {
if (board[i][col] == key) return false;
}
// 检查四方格
int rows = row / 3;
int cols = col / 3;
for (int i = 3 * rows; i < 3 * rows + 3; i++) {
for (int j = 3 * cols; j < 3 * cols + 3; j++) {
if (board[i][j] == key) return false;
}
}
return true;
}
void solveSudoku(vector<vector<char>>& board) {
backtracking(board);
return;
}
};二、题目思路整理
原文地址：https://blog.csdn.net/CXDNW/article/details/142694444
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