LeetCode - N-Queens I && II

N-Queens I

https://leetcode.com/problems/n-queens/

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 two distinct solutions to the 4-queens puzzle:

[
[".Q..",  // Solution 1
"...Q",
"Q...",
"..Q."],

["..Q.",  // Solution 2
"Q...",
"...Q",
".Q.."]
]

这道题跟solve sudoku差不多，都是每个index挨个检查是否valid，如果valid就进入下一轮递归，否则就继续寻找。
注意这里，用一个LinkedList<Integer>来存已经生成的position，list中第i个元素j表示第i行的queen在第j列。这样简化了很多代码，不然直接操作string[]简直是灾难啊。

还有就是在检查对角线时，检查的是 Math.abs(row1-row2)==Math.abs(col1-col2)，如果检查Math.abs(row1-col) == Math.abs(row2-col2)的话，只能检查一个方向的对象线，即从左上到右下的，不能检查从右上到左下的对角线。

从左上到右下的对角线上row-col值是相同的

从右上到左下的对角线row+col的值相同

两种都满足 Math.abs(row1-row2)==Math.abs(col1-col2)

public class Solution {
public List<String[]> solveNQueens(int n) {
List<String[]> rst = new LinkedList<String[]>();
List<Integer> position = new LinkedList<Integer>();

helper(rst, position, n);
return rst;
}

public void helper(List<String[]> rst, List<Integer> position, int n){
if(position.size()==n){
genBoard(position, rst);
return;
}
for(int i=0; i<n; i++){
if(isValid(position, i)){
position.add(i);
helper(rst, position, n);
position.remove(position.size()-1);
}
}
}

public boolean isValid(List<Integer> position, int col){
int row = position.size();
for(int i=0; i<position.size(); i++){
if(col == position.get(i)) return false;
if(Math.abs(row-i)==Math.abs(col-position.get(i))) return false;
}
return true;
}

public void genBoard(List<Integer> position, List<String[]> rst){
String[] board = new String[position.size()];
for(int i=0; i<position.size(); i++){
int queen = position.get(i);
StringBuilder sb = new StringBuilder();
for(int j=0; j<position.size(); j++){
if(j!=queen) sb.append('.');
else sb.append('Q');
}
board[i] = sb.toString();
}
rst.add(board);
}
}

N-Queen II

https://leetcode.com/problems/n-queens-ii/

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.



这道题直接用上一道题的函数就可以了，只是当遇到一个solution的时候，不需要生成board，只需要count++就可以了。另外，由于JAVA都是值传递，所以如果
传给helper函数的是int count的话，那么helper中修改的count值不会生效，所以这里传一个int[] count数组，然后修改count[0]，另一个方法是在类中设置一个全局变量。

public class Solution {
public int totalNQueens(int n) {
List<String[]> rst = new LinkedList<String[]>();
List<Integer> position = new LinkedList<Integer>();
int[] count = new int[1];

helper(rst, position, n, count);
return count[0];
}

public void helper(List<String[]> rst, List<Integer> position, int n, int[] count){
if(position.size()==n){
count[0]++;
return;
}
for(int i=0; i<n; i++){
if(isValid(position, i)){
position.add(i);
helper(rst, position, n, count);
position.remove(position.size()-1);
}
}
}

public boolean isValid(List<Integer> position, int col){
int row = position.size();
for(int i=0; i<position.size(); i++){
if(col == position.get(i)) return false;
if(Math.abs(row-i)==Math.abs(col-position.get(i))) return false;
}
return true;
}
}