LeetCode233 Number of Digit One

本文介绍了一个算法问题,即计算从0到n的所有非负整数中数字1出现的总次数。通过一个Java实现的例子,展示了如何高效地解决这个问题。示例中包括了对13作为输入的解释,以及一个对大规模数字1410065408进行计算的演示。

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题目:

       Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

       Example:

       Input: 13

       Output: 6

       Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

public class NumberDigitOne {
    public int countDigitOne(int n) {
        int count = 0, previous = 0, coef = 1;
        while (n > 0) {
            int remain = n % 10;
            int over = n / 10;
            if (remain > 1) {
                count += coef;
            } else if (remain == 1) {
                count += previous + 1;
            }
            count += coef * over;
            previous += coef * remain;
            coef *= 10;
            n /= 10;
        }
        return count;
    }

    public static void main(String[] args) {
        int n = 1410065408;
        System.out.println(new NumberDigitOne().countDigitOne(n));
    }
}

 

 

 

### LeetCode Problem 37: Sudoku Solver #### Problem Description The task involves solving a partially filled Sudoku puzzle. The input is represented as a two-dimensional integer array `board` where each element can be either a digit from '1' to '9' or '.' indicating empty cells. #### Solution Approach To solve this problem, one approach uses backtracking combined with depth-first search (DFS). This method tries placing numbers between 1 and 9 into every cell that contains '.', checking whether it leads to a valid solution by ensuring no conflicts arise within rows, columns, and subgrids[^6]. ```cpp void solveSudoku(vector<vector<char>>& board) { backtrack(board); } bool backtrack(vector<vector<char>> &board){ for(int row = 0; row < 9; ++row){ for(int col = 0; col < 9; ++col){ if(board[row][col] != '.') continue; for(char num='1';num<='9';++num){ if(isValidPlacement(board,row,col,num)){ placeNumber(num,board,row,col); if(backtrack(board)) return true; removeNumber(num,board,row,col); } } return false; } } return true; } ``` In the provided code snippet: - A function named `solveSudoku()` initiates the process. - Within `backtrack()`, nested loops iterate over all positions in the grid looking for unassigned spots denoted by '.' - For any such spot found, attempts are made to insert digits ranging from '1' through '9'. - Before insertion, validation checks (`isValidPlacement`) ensure compliance with Sudoku rules regarding uniqueness per row/column/subgrid constraints. - If inserting a number results in reaching a dead end without finding a complete solution, removal occurs before trying another possibility. This algorithm continues until filling out the entire board correctly or exhausting possibilities when returning failure status upward along recursive calls stack frames. --related questions-- 1. How does constraint propagation improve efficiency while solving puzzles like Sudoku? 2. Can genetic algorithms provide alternative methods for tackling similar combinatorial problems effectively? 3. What optimizations could enhance performance further beyond basic DFS/backtracking techniques used here?
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