zoukankan      html  css  js  c++  java
  • [LeetCode] N-Queens II

    If you have solved the N-Queens problem, this one can be solved in a similar manner. Starting from the first row, we try each of its columns. If there is no attack, we move on to the next row based previous rows. Otherwise, we backtrack to the current row and try another selection of column position. Once we meet the last row, increase the counts by 1.

    The code is as follows.

     1 class Solution {
     2 public:
     3     int totalNQueens(int n) {
     4         int* colPos = new int [n];
     5         int counts = 0;
     6         solve(colPos, n, 0, counts);
     7         delete colPos;
     8         return counts;
     9     }
    10 private:
    11     bool noAttack(int*& colPos, int row, int col) {
    12         for (int r = row - 1, ld = col - 1, rd = col + 1; r >= 0; r--, ld--, rd++)
    13             if (colPos[r] == col || colPos[r] == ld || colPos[r] == rd)
    14                 return false;
    15         return true;
    16     }
    17     void solve(int*& colPos, int n, int row, int& counts) {
    18         if (row == n) {
    19             counts++;
    20             return;
    21         }
    22         for (int col = 0; col < n; col++) {
    23             colPos[row] = col;
    24             if (noAttack(colPos, row, col))
    25                 solve(colPos, n, row + 1, counts);
    26         }
    27     }
    28 };

    Someone has even suggested a damn clever solution to this problem using bit-manipulations in this link (refer to the first answer).

    I rewrite the code below for reference.

     1 class Solution {
     2 public:
     3     int totalNQueens(int n) {
     4         int counts = 0;
     5         int limit = (1 << n) - 1;
     6         solve(0, 0, 0, limit, counts);
     7         return counts;
     8     }
     9 private:
    10     void solve(int hProj, int lProj, int rProj, int limit, int& counts) {
    11         if (hProj == limit) {
    12             counts++;
    13             return;
    14         }
    15         int pos = limit & (~(hProj | lProj | rProj));
    16         while (pos) {
    17             int p = pos & (-pos);
    18             pos ^= p;
    19             solve(hProj | p, (lProj | p) << 1, (rProj | p) >> 1, limit, counts);
    20         }
    21     }
    22 };

    The above two are both recursive solutions. This problem can also be solved iteratively as in this link.

    I have also rewritten the code below for reference.

     1 class Solution {
     2 public:
     3     int totalNQueens(int n) {
     4         int* colPos = new int [n];
     5         memset(colPos, -1, n * sizeof(int));
     6         int row = 0, counts = 0;
     7         while (row >= 0) {
     8             if (row == n) {
     9                 counts++;
    10                 row--;
    11             }
    12             int col = (colPos[row] == -1) ? 0 : colPos[row] + 1;
    13             for (; col < n; col++) {
    14                 if (noAttack(colPos, row, col)) {
    15                     colPos[row] = col;
    16                     row++;
    17                     break;
    18                 }
    19             }
    20             if (col == n) {
    21                 colPos[row] = -1;
    22                 row--;
    23             }
    24         }
    25         return counts;
    26     }
    27 private:
    28     bool noAttack(int*& colPos, int row, int col) {
    29         for (int r = row - 1, ld = col - 1, rd = col + 1; r >= 0; r--, ld--, rd++)
    30             if (colPos[r] == col || colPos[r] == ld || colPos[r] == rd)
    31                 return false;
    32         return true;
    33     }
    34 };
  • 相关阅读:
    linux性能查看调优
    免密登录
    nginx配置
    Samba
    硬RAID与软RAID的区别
    LVM-扩容目录
    解决表面磁盘满,而实际没有大文件的问题
    LINUX下的JENKINS+TOMCAT+MAVEN+GIT+SHELL环境的搭建使用(JENKINS自动化部署)
    Docker 容器使用
    docker基础
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4614716.html
Copyright © 2011-2022 走看看