题目出处
题目描述
个人解法
思路：
todo
代码示例：（Java）
todo
复杂度分析
todo
官方解法
前言
方法1：基于集合的回溯
思路：
代码示例：（Java）
public class Solution1 {
public int totalNQueens(int n) {
Set<Integer> columns = new HashSet<Integer>();
Set<Integer> diagonals1 = new HashSet<Integer>();
Set<Integer> diagonals2 = new HashSet<Integer>();
return backtrack(n, 0, columns, diagonals1, diagonals2);
}
public int backtrack(int n, int row, Set<Integer> columns, Set<Integer> diagonals1, Set<Integer> diagonals2) {
if (row == n) {
return 1;
} else {
int count = 0;
for (int i = 0; i < n; i++) {
if (columns.contains(i)) {
continue;
}
int diagonal1 = row - i;
if (diagonals1.contains(diagonal1)) {
continue;
}
int diagonal2 = row + i;
if (diagonals2.contains(diagonal2)) {
continue;
}
columns.add(i);
diagonals1.add(diagonal1);
diagonals2.add(diagonal2);
count += backtrack(n, row + 1, columns, diagonals1, diagonals2);
columns.remove(i);
diagonals1.remove(diagonal1);
diagonals2.remove(diagonal2);
}
return count;
}
}
}
复杂度分析
方法2:基于位运算的回溯
思路：
代码示例：（Java）
public class Solution2 {
public int totalNQueens(int n) {
return solve(n, 0, 0, 0, 0);
}
public int solve(int n, int row, int columns, int diagonals1, int diagonals2) {
if (row == n) {
return 1;
} else {
int count = 0;
int availablePositions = ((1 << n) - 1) & (~(columns | diagonals1 | diagonals2));
while (availablePositions != 0) {
int position = availablePositions & (-availablePositions);
availablePositions = availablePositions & (availablePositions - 1);
count += solve(n, row + 1, columns | position, (diagonals1 | position) << 1, (diagonals2 | position) >> 1);
}
return count;
}
}
}
复杂度分析
小结
