Zero Must-Read Series 4 Backtracking algorithm problem solving routine framework

table of Contents

1. Full arrangement problem

Second, the N queen problem

Three, summary

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The backtracking problem is actually a decision tree traversal problem. Just think about 3 questions:

• Path: the choice made;
• Choice list: the choices currently available;
• End condition: A condition that can no longer be selected.

frame:

result = []
def backtrack(路径, 选择列表):
    if 满足结束条件:
        result.add(路径)
        return
​
    for 选择 in 选择列表:
        做选择
        backtrack(路径, 选择列表)
        撤销选择

There is an operation to make selection and undo selection, namely add and then remove.

1. Full arrangement problem

Get all the paths in the full arrangement. Assuming that n numbers are not the same, there are n in total! Paths.

List<List<Integer>> res = new LinkedList<>();
​
/* 主函数，输入一组不重复的数字，返回它们的全排列 */
List<List<Integer>> permute(int[] nums) {
    // 记录「路径」
    LinkedList<Integer> track = new LinkedList<>();
    backtrack(nums, track);
    return res;
}
​
// 路径：记录在 track 中
// 选择列表：nums 中不存在于 track 的那些元素
// 结束条件：nums 中的元素全都在 track 中出现
void backtrack(int[] nums, LinkedList<Integer> track) {
    // 触发结束条件
    if (track.size() == nums.length) {
        res.add(new LinkedList(track));
        return;
    }
​
    for (int i = 0; i < nums.length; i++) {
        // 排除不合法的选择
        if (track.contains(nums[i]))
            continue;
        // 做选择
        track.add(nums[i]);
        // 进入下一层决策树
        backtrack(nums, track);
        // 取消选择
        track.removeLast();
    }
}

Second, the N queen problem

Give you an N×N chessboard and let you place N queens so that they cannot attack each other. The queen can attack any unit in the same row, same column, upper left, lower left, upper right, and lower right.

vector<vector<string>> res;
​
/* 输入棋盘边长 n，返回所有合法的放置 */
vector<vector<string>> solveNQueens(int n) {
    // '.' 表示空，'Q' 表示皇后，初始化空棋盘。
    vector<string> board(n, string(n, '.'));
    backtrack(board, 0);
    return res;
}
​
// 路径：board 中小于 row 的那些行都已经成功放置了皇后
// 选择列表：第 row 行的所有列都是放置皇后的选择
// 结束条件：row 超过 board 的最后一行
void backtrack(vector<string>& board, int row) {
    // 触发结束条件
    if (row == board.size()) {
        res.push_back(board);
        return;
    }
​
    int n = board[row].size();
    for (int col = 0; col < n; col++) {
        // 排除不合法选择
        if (!isValid(board, row, col)) 
            continue;
        // 做选择
        board[row][col] = 'Q';
        // 进入下一行决策
        backtrack(board, row + 1);
        // 撤销选择
        board[row][col] = '.';
    }
}

The isValid function is to exclude illegal choices:

/* 是否可以在 board[row][col] 放置皇后？ */
bool isValid(vector<string>& board, int row, int col) {
    int n = board.size();
    // 检查列是否有皇后互相冲突
    for (int i = 0; i < n; i++) {
        if (board[i][col] == 'Q')
            return false;
    }
    // 检查右上方是否有皇后互相冲突
    for (int i = row - 1, j = col + 1; 
            i >= 0 && j < n; i--, j++) {
        if (board[i][j] == 'Q')
            return false;
    }
    // 检查左上方是否有皇后互相冲突
    for (int i = row - 1, j = col - 1;
            i >= 0 && j >= 0; i--, j--) {
        if (board[i][j] == 'Q')
            return false;
    }
    return true;
}

If you find an answer, just return, you can modify it to return early:

// 函数找到一个答案后就返回 true
bool backtrack(vector<string>& board, int row) {
    // 触发结束条件
    if (row == board.size()) {
        res.push_back(board);
        return true;
    }
    ...
    for (int col = 0; col < n; col++) {
        ...
        board[row][col] = 'Q';
​
        if (backtrack(board, row + 1))
            return true;
​
        board[row][col] = '.';
    }
​
    return false;
}

Three, summary

Streamline the core framework:

def backtrack(...):
    for 选择 in 选择列表:
        做选择
        backtrack(...)
        撤销选择