[Leetcode 51&52] N-queens I & II (n Queen's question)

Title Link: N-queens

Import Java.util.arraylist;import java.util.arrays;import java.util.list;/** * The N-queens puzzle is the problem of Placi ng N Queens on a nxn chessboard such that no, and Queens attack each other. Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains a distinct board configuration of the N-queens ' placement, where ' Q ' and '. ' Both indicate a queen and an empty space respectively. For example,there exist, distinct solutions to the 4-queens puzzle:[[". Q.. ",//solution 1" ... Q "," Q ... ",".. Q. "], ["]. Q. ",//Solution 2" Q ... "," ... Q ",". Q.. "] * */public class Nqueens {//9/9 test Cases passed.//status:accepted//runtime:217 ms//submitted:0 minutes ago//backtracking//Time  Complexity O (n!) space complexity O (n) public int[] columns;//store placed Queens occupy columns, 0 not occupied, 1 occupied public int[] main_diag;//store placed Queens occupy the main diagonal, 0 is not occupied, 1 is occupied by public Int[] anti_diag;//Store placed Queens occupy the diagonal, 0 not occupied, 1 occupied public string[] st_str;public list<string[]> nqueens = new Arraylist&lt ; String[]> ();p ublic void Init (inT n) {columns = new int[n];        Main_diag = new int[2 * n];        Anti_diag = new int[2 * n];        Arrays.fill (columns, 0);        Arrays.fill (main_diag, 0);    Arrays.fill (anti_diag, 0); Createstring (n);} public void createstring (int n) {st_str = new string[n]; StringBuilder sb = new StringBuilder (); for (int i = 0; i < n; i++) {Sb.append (".");}    for (int i = 0; i < n; i++) {Sb.replace (i, i + 1, "Q"); St_str[i] = sb.tostring (); Sb.replace (i, i + 1, ".");}}        Public list<string[]> solvenqueens (int n) {//Initialize each state variable init (n);    int[] C = new Int[n];        Arrays.fill (C, 0);    DFS (C, 0);    return nqueens;    } public void Dfs (int[] C, int row) {int N = c.length;    means to find a viable solution if (row = = N) {string[] Nqueen = new String[n];    for (int i = 0; i < N; i++) {nqueen[i] = St_str[c[i]];}    Nqueens.add (Nqueen);    Return } for (int i = 0; i < N; i++) {//If not legal, this skips the current loop if (! ( Columns[i] = = 0&& Main_diag[i + row] = = 0&& Anti_diag[n-i + row] = = 0)) continue; Execute C[row] = i;columns[i] = 1;main_diag[i + row] = 1;anti_diag[n-i + row] = 1;dfs (C, row + 1);//undo C[row] = 0;columns[i]       = 0;main_diag[i + Row] = 0;anti_diag[n-i + row] = 0;} } public static void Main (string[] args) {Nqueens queens = new Nqueens (); list<string[]> result = Queens.solvenqueens (8); System.out.println (Result.size ());//for (int i = 0; i < result.size (); i++) {//for (int j = 0; J < Result.get (i). Len Gth J + +) {//system.out.println (Result.get (i) [j]);//}//system.out.println ();/}}}

Title Link: n-queens-ii

Import java.util.arrays;/** * Follow up for n-queens problem. Now, instead outputting board configurations and return the total number of distinct solutions. * */public class Nqueensii {//9/9 test Cases passed.//status:accepted//runtime:231 ms//submitted:0 minutes ago//backtracking// Time complexity O (n!) space complexity O (n) public int[] columns;//store placed Queens occupy columns, 0 not occupied, 1 occupied public int[] main_diag;//store placed Queens occupy the main diagonal, 0 is not accounted for, 1 is accounted for publ        IC int[] anti_diag;//store placed Queens occupy the diagonal, 0 not accounted for, 1 occupied public int total;public void init (int n) {columns = new int[n];        Main_diag = new int[2 * n];        Anti_diag = new int[2 * n];        Arrays.fill (columns, 0);        Arrays.fill (main_diag, 0);           Arrays.fill (anti_diag, 0);        Total = 0;//count} public int totalnqueens (int n) {//Initialize each state variable init (n);    int[] C = new Int[n];        Arrays.fill (C, 0);    DFS (C, 0);    return total;    } public void Dfs (int[] C, int row) {int N = c.length;    means to find a viable solution if (row = = N) {total + +;    Return } foR (int i = 0; i < N; i++) {//If not legal, this skips the current loop if (! (    Columns[i] = = 0&& Main_diag[i + row] = = 0&& Anti_diag[n-i + row] = = 0)) continue; Execute C[row] = i;columns[i] = 1;main_diag[i + row] = 1;anti_diag[n-i + row] = 1;dfs (C, row + 1);//undo C[row] = 0;columns[i]       = 0;main_diag[i + Row] = 0;anti_diag[n-i + row] = 0;} public static void Main (string[] args) {nqueensii queens = new Nqueensii (), for (int i = 0; i <; i++) {Syste M.OUT.PRINTLN (i + "-Queen has" + queens.totalnqueens (i) + "solution");}}} The Queen of output//0-has 1 solutions//1-Queen has 1 solutions//2-Queen has 0 solutions//3-Queen has 0 solutions//4-Queen has 2 solutions//5-Queen has 10 solutions//6-Queen has 4 solutions//7-Queen has 40 kinds of solution 8-Queens have 92 solutions//9-Queen has 352 solutions//10-Queen has 724 solutions//11-Queen has 2680 methods//12-Queen has 14200 solutions//13-Queen has 73712 solutions//14-Queen 365596 kinds Solution//15-Queen has 2279184 kinds of solution

[Leetcode 51&52] N-queens I & II (n Queen's question)