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
一只不想敲代码的程序猿1 分钟前
如何更高效地拆解问题?
算法
微尘hjx1 小时前
【Deep-OC-SORT 02】多轨迹算法Deep-OC-SORT-相对 OC-SORT 的改进点
人工智能·算法·yolo·目标跟踪·视觉检测·ultralytics·deep-oc-sort
梦帮科技1 小时前
意识启发式量子隧穿退火算法用于全局优化 (CITA) 完整技术指南
网络·人工智能·算法·量子计算·模拟退火算法
一去不复返的通信er2 小时前
排序 + 双指针:力扣15三数之和
算法·leetcode
文丑颜不良啊2 小时前
LeetCode1901~1920题目解析
java·leetcode
Pniubi3 小时前
力扣3无重复字符的最长字串(滑动窗口)
算法·leetcode·职场和发展
On The Road :3 小时前
深入理解 JVM 垃圾回收:GC 是什么、如何判定垃圾、四大回收算法全解析
jvm·算法
智者知已应修善业3 小时前
【51单片机实现从下到上,从上到下,从中间到两边,两边到中间】2024-10-23
c语言·经验分享·笔记·单片机·嵌入式硬件·51单片机
微尘hjx4 小时前
【Deep-OC-SORT 03】多轨迹算法Deep-OC-SORT-卡尔曼滤波预测过程
算法·yolo·目标跟踪·ultralytics·车跟踪·轨迹算法·deep-oc-sort
老师我太想进步了20266 小时前
【数据结构-3】算法美学之降复杂度
数据结构·算法