LeetCode //C - 52. N-Queens II

52. N-Queens II

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

Given an integer n , return the number of distinct solutions to the n-queens puzzle.

Example 1:

Input: n = 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown.

Example 2:

Input: n = 1
Output: 1

Constraints:
  • 1 <= n <= 9

From: LeetCode

Link: 52. N-Queens II


Solution:

Ideas:
  1. Start in the leftmost column.
  2. If all queens are placed, return true.
  3. Try all rows in the current column.
  • If the queen can be placed safely in this row, mark this cell and place the queen.
  • Recur to place the rest of the queens.
  • If placing the queen in the current row and proceeding to place the next queen leads to a solution, return true.
  • If placing the queen doesn't lead to a solution, then unmark this cell, backtrack, and go to the next row in the current column.
  1. If all rows have been tried and none worked, return false to trigger backtracking.

To determine if a queen can be placed safely, we need to check three things:

  1. There's no queen in the same row.
  2. There's no queen in the same diagonal (both left upper diagonal and right upper diagonal).
Code:
c 复制代码
bool isSafe(int board[], int row, int col, int n) {
    for (int i = 0; i < col; i++) {
        if (board[i] == row || 
            board[i] - i == row - col || 
            board[i] + i == row + col) {
            return false;
        }
    }
    return true;
}

int solveNQueensUtil(int n, int col, int board[]) {
    if (col >= n) {
        return 1;
    }
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (isSafe(board, i, col, n)) {
            board[col] = i;
            count += solveNQueensUtil(n, col + 1, board);
        }
    }
    return count;
}

int totalNQueens(int n) {
    int board[n];
    for (int i = 0; i < n; i++) {
        board[i] = 0;
    }
    return solveNQueensUtil(n, 0, board);
}
相关推荐
学逆向的5 小时前
C 枚举变量
c语言·开发语言·网络安全
冷凝雨5 小时前
前导零、后导零、最高有效位、最低有效位计算(C)
c语言
鸥梨菌Honevid6 小时前
C 语言 restrict 关键字深度解析:从“独占访问”承诺到编译器激进优化
c语言
企业数字化笔记7 小时前
视频人体姿态怎么做三维可视化?2D关键点、深度局限与技术选型
人工智能·算法·音视频
?abc!8 小时前
(数组)581. 最短无序连续子数组
数据结构·算法·leetcode
小郎君65811 小时前
MPTT树复选框的半选状态:一棵百万节点树从递归到计数器,再到区间覆盖
算法
酌沧11 小时前
MemT让记忆成为模型内生能力
jvm·人工智能·算法
水饺编程11 小时前
第2章:Linux 基础,切换用户
linux·c语言·汇编
hold?fish:palm11 小时前
40 二叉树的直径
c++·leetcode
Shimeri11 小时前
越卷越像:我用遗传算法做了个"内卷"实验
算法·three.js·数据可视化·遗传算法·进化算法