Leetcode N-Queens问题

题目描述：

The n-queens puzzle is the problem of placing n queens on an n×n
chessboard such that no two 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.

解决方法：
两个方法都是通过递归来解决问题，方法一是我自己编写的，方法二是参考leetcode代码的，方法二的思路与方法一类似，但是方法二对皇后状态的存储进行优化，加快了运行速度同时也减少了运行时间。

class Solution {
public List<List<String>> solveNQueens(int n) {
int[][] chessboard = new int[n][n];
int[][] au = new int[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
au[i][j] = n;
return solve(chessboard, au, 0, n);
}

public List<List<String>> solve(int[][] chessboard, int[][] au, int row, int n){
List<List<String>> res = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++){
sb.append(".");
}

if (row == n-1){
for (int i = 0; i < n; i++){
if (chessboard[row][i] == 0){
List<String> temp = new ArrayList<>();
sb.replace(i,i+1,"Q");
temp.add(sb.toString());
res.add(temp);
return res;
}
}
}

for (int i = 0; i < n; i++){
if (chessboard[row][i] == 0){
chessboard[row][i] = 1;
au[row][i] = Math.min(row,au[row][i]);
sb.replace(i,i+1,"Q");
for (int j = row+1; j < n; j++){
chessboard[j][i] = 1;
au[j][i] = Math.min(row,au[j][i]);
if ((j-row+i)<= n-1){
chessboard[j][j-row+i] = 1;
au[j][j-row+i] = Math.min(au[j][j - row + i], row);
}
if ((i-j+row) >= 0){
chessboard[j][i-j+row] = 1;
au[j][i-j+row] = Math.min(au[j][i - j + row], row);
}
}
List<List<String>> l = solve(chessboard, au, row+1, n);
for (List<String> strings : l) {
strings.add(0, sb.toString());
res.add(strings);
}
sb.replace(i,i+1,".");
if (au[row][i] >= row){
chessboard[row][i] = 0;
au[row][i] = n;
}
for (int j = row+1; j < n; j++){
if (au[j][i] >= row && au[j][i] >= row){
chessboard[j][i] = 0;
au[j][i] = n;
}
if ((j-row+i)<= n-1 && au[j][j-row+i]>=row){
chessboard[j][j-row+i] = 0;
au[j][j-row+i] = n;
}
if ((i-j+row) >= 0 && au[j][i-j+row]>=row){
chessboard[j][i-j+row] = 0;
au[j][i-j+row] = n;
}
}
}
}
return res;
}
}

上叙方法的不足之处对每一行每一列皇后状态的存储，通过int来存储浪费操作的时间和空间，通过位来进行存储可以极大的节约时间

int availableBit = (~usedBits) & (usedBits + 1);

通过该方法可以获取到最后一个可以用的位，就是可以被赋值的位置，使用位操作还有一个好处就是不需要恢复之前的状态

usedBits |= availableBit;

该方法可以给特定的位进行赋值。

以下是参考leetcode的代码

class Solution {
public List<List<String>> solveNQueens(int n) {
List<List<String>> results = new ArrayList();
char[] letters = new char[n];
Arrays.fill(letters, '.');
int[] positions = new int[n];
find(0, 0, 0, 0, n, results, letters, positions);
return results;
}

private void find(int index, int upBits, int leftBits, int rightBits, int size, List<List<String>> results, char[] letters, int[] positions) {
int usedBits = upBits | leftBits | rightBits;

for (int i = 0; i < size; ++i) {
int availableBit = (~usedBits) & (usedBits + 1);
if (availableBit == (1 << i)) {
positions[index] = i;
usedBits |= availableBit;
if (index == size - 1) {
List<String> result = new ArrayList();

for (int position : positions) {
letters[position] = 'Q';
result.add(new String(letters));
letters[position] = '.';
}

results.add(result);
} else {
find(index + 1,
upBits | availableBit,
(leftBits | availableBit) << 1,
(rightBits | availableBit) >> 1,
size,
results,
letters,
positions);
}
}
}
}
}