2024.06.24 刷题日记

17. 电话号码的字母组合

依然是昨天的回溯，思路是根据 index，来确定要回溯的对象：

class Solution {
public:vector<string> letterCombinations(string digits) {vector<string> results;if (digits.empty())return results; // 处理空输入// 数字到字母的映射vector<string> posi = {"",    "",    "abc",  "def", "ghi","jkl", "mno", "pqrs", "tuv", "wxyz"};string current; // 当前正在构建的字符串backtrack(digits, 0, current, posi, results);return results;}void backtrack(const string& digits, int index, string& current,const vector<string>& posi, vector<string>& results) {if (digits.size() == current.size()) {results.push_back(current);return;}int pos = digits[index] - '0';string letters = posi[pos];for (auto letter : letters) {current.push_back(letter);backtrack(digits, index + 1, current, posi, results);current.pop_back();}}
};

22. 括号生成

这道题目挺难，这里的思路是利用两个规则来构建合法的括号表达式：

		if (open < max) {current.push_back('(');}if (close < open) {current.push_back(')');}

如果左括号小于 max，就加，如果右括号小于左括号，就加。这两个规则可以确保生成的括号表达式是合法的。因为这在根本上就规避了这样的组合：)((...、())...：

class Solution {
public:vector<string> generateParenthesis(int n) {vector<string> result;string current;backtrack(result, current, 0, 0, n);return result;}void backtrack(vector<string>& result, string& current, int open, int close,int max) {if (current.size() == max * 2) {result.push_back(current);return;}if (open < max) {current.push_back('(');backtrack(result, current, open + 1, close, max);current.pop_back();}if (close < open) {current.push_back(')');backtrack(result, current, open, close + 1, max);current.pop_back();}}
};

79. 单词搜索

这个题目还是回溯，需要一个 helper 数组来标记已经访问过的位置，并且回溯过程中，递归函数也是带有返回值的。在调用函数中，需要对每个位置进行搜索，如果搜索到就返回，相当于示例优化；在被调用函数中，如果board[x][y] != word[index]就返回 false，相当于剪枝，因此整个过程时间复杂度并不高：

class Solution {
public:bool exist(vector<vector<char>>& board, string word) {int m = board.size(), n = board[0].size();vector<vector<bool>> visited(m, vector<bool>(n, false));for (int i = 0; i < m; ++i) {for (int j = 0; j < n; ++j) {if (search(board, word, 0, i, j, visited))return true;}}return false;}private:bool search(vector<vector<char>>& board, string& word, int index, int x,int y, vector<vector<bool>>& visited) {if (index == word.size())return true;if (x < 0 || x >= board.size() || y < 0 || y >= board[0].size() ||visited[x][y] || board[x][y] != word[index])return false;visited[x][y] = true;bool found = search(board, word, index + 1, x + 1, y, visited) ||search(board, word, index + 1, x - 1, y, visited) ||search(board, word, index + 1, x, y + 1, visited) ||search(board, word, index + 1, x, y - 1, visited);visited[x][y] = false;return found;}
};

131. 分割回文串

这个题目的核心思路是剪枝：当 (s,index,i) 是回文串的时候，纵向深入：

class Solution {
public:vector<vector<string>> partition(string s) {vector<vector<string>> result;vector<string> path;backtrace(s, 0, path, result);return result;}private:void backtrace(string& s, int index, vector<string>& path,vector<vector<string>>& result) {if (index == s.size()) {result.push_back(path);return;}for (int i = index; i < s.size(); i++) {if (isPalindrome(s, index, i)) {path.push_back(s.substr(index, i - index + 1));backtrace(s, i + 1, path, result);path.pop_back();}}}bool isPalindrome(string& str, int start, int end) {while (start < end) {if (str[start] != str[end]) {return false;}start++;end--;}return true;}
};

51. N 皇后

N 皇后相关的题目很多，但是都大同小异。这个题目的难点在于如果去标记纵向、两个对角线是否被占用，对于第 row 行、col 列，那么对角线为：diag1[row - col + n - 1]、diag2[row + col]。其中：

vector<bool> diag1(2 * n - 1, false); // 左下到右上
vector<bool> diag2(2 * n - 1, false); // 左上到右下

最好把这个记住，可以加快做题速度。

解决了这个问题之后，其他就很简单了，在 row 上进行回溯就行了：

class Solution {
public:vector<vector<string>> solveNQueens(int n) {vector<vector<string>> solutions;vector<string> board(n, string(n, '.'));vector<bool> cols(n, false);vector<bool> diag1(2 * n - 1, false);vector<bool> diag2(2 * n - 1, false);backtrack(solutions, board, cols, diag1, diag2, 0, n);return solutions;}private:void backtrack(vector<vector<string>>& solutions, vector<string>& board,vector<bool>& cols, vector<bool>& diag1, vector<bool>& diag2,int row, int n) {if (row == n) {solutions.push_back(board);return;}for (int col = 0; col < n; ++col) {if (cols[col] || diag1[row - col + n - 1] || diag2[row + col]) {continue;}board[row][col] = 'Q';cols[col] = diag1[row - col + n - 1] = diag2[row + col] = true;backtrack(solutions, board, cols, diag1, diag2, row + 1, n);board[row][col] = '.';cols[col] = diag1[row - col + n - 1] = diag2[row + col] = false;}}
};