# Leetcode 51.n-queens (N queen problem) thought and method of solving problems

Source: Internet
Author: User

N-queens

The N-queens puzzle is the problem of placing n queens in an 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.. "]

]

Idea: The big way I think of two ideas, the first one is to transform the N queen problem into a full permutation problem. Set X[n] As a set of solutions, the Queen is placed on behalf of row I, section x[i] column.

So find out the whole arrangement and then judge whether it is legal.

The code is as follows (without AC, which should be AC after further optimization):

`public class Solution {public list<list<string>> solvenqueens (int n) {LIST&LT;LIST&LT;STRING&GT;&G T    returnlist = new arraylist<list<string>> ();        /** * X[i] for the location of column I * converted to 0-(n-1) of the full arrangement, and then determine whether the full alignment of the position is valid */int[] num = new Int[n];        int i = 0;        while (I < n) {num[i] = i;//padding array is 0-(n-1) i++;        } list<list<integer>> List = Permuteunique (num); for (list<integer> al:list) {//per group of data processing for (i = 0; i < al.size (); i++) {if (!check (Al, I, Al.get (i)) break;//not valid out} if (i = = N) {///to last, stating all legal list<string> ls = new Arraylist<string&gt        ;(); for (i = 0; i < al.size (); i++) {string s = "";//String for per group (int j = 0; J < n;j++) {if (j = = Al.get (i)) {s + = "Q";//Place Queen's position} else{s + = "."; /Place}} ls.add (s);//Add to LS} returnlist.add (LS);//Add a set of solutions       }} return returnlist; }/** * Determine if it is legal * @param al a set of solutions * @param i current row * @param x Current column * @return Legal */boolean check (list<integer> al,int i,int x) {for (int k = 0; k < i; k++) {//only compared with above, otherwise if (Al.get (k) = = X | | x-i = = al.get (k)-K | | x + i = = Al.get (k)    + K) return false;    }return true; }//Perfection Arrange public static list<list<integer>> permuteunique (int[] num) {list<list<integer>> returnlist = new arraylist<list<integer>> (); Returnlist.add (new arraylist<integer> ()); for (int i = 0; i < num.length; i++) {set<list<integer>> currentset = new Hashset<list<integer>&gt ;(); for (list<integer> l:returnlist) {for (int j = 0; J < l.size () + 1; j + +) {L.add (J, Num[i]); list<integer> T = new arraylist<integer> (l); L.remove (j); Currentset.add (t);}} returnlist = new arraylist<list<integer>> (currentset);} return returnlist;}}`

Another way of thinking is backtracking, in line with the requirements of the go down, do not meet the requirements of the backward.

Specific code:

`public class Solution {/** * The whole idea of this question is to build an array of x[n], meaning the first place in the X[i] column to place the Queen's legal judgment rule is the column is not equal, the slash is not equal * i + x[i] = = j + X[J] denotes a forward slash consistent * X[i]-i = x[j]-J means that a negative slash is not consistent on the legal */list<list<string>> list;//save result public list<list <String>> solvenqueens (int n) {list = new arraylist<list<string>> (); int[] x = new int[n];//save result Quee NS (x, n, 0); return list;} void Queens (int[] x,int n,int row) {for (int i = 0; i < n; i++) {if (check (x,n,row,i)) {//Judge legal x[row] = i;//Place queen on row row, column I if ( row = = n-1) {///If it is the last row, the output addlist (x,n); X[row] = 0;//backtracking, looking for the next result return;} Queens (x, N, row+1);//Look for the next line x[row] = 0;//back}}}/** * Add the results of each group to the list * @param x Array Solution * @param n checkerboard width */private void addlist (int [] x,int N) {//Add result string[][] SArr = new String[n][n]; List<string> al = new arraylist<string> (); for (int i = 0; i < n; i++) {Arrays.fill (Sarr[i], "."); /fill in. sarr[i][x[i]] = "Q";//position qstring s = ""; for (String Str:sarr[i]) {s + = str;//add together}al.add (s);//Add one row}list.add (AL) ;//Add a set of solutions}/** * @param x Array Solution * @param n Checkerboard Width * @param index currently placed row * @param i currently place column * @return */boolean check (int[] x,int n,int row, int Col) {for (int i = 0; i < row; i++) {//Because the rows are not equal, determine if the columns are equal, if the slashes are equal if (x[i] = Col | | x[i] + i = col + row | | x[i]-i = = C Ol-row) return false;} return true;}}`

Leetcode 51.n-queens (N queen problem) thought and method of solving problems

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.