算法 中等 | 34. N皇后问题 II
题目描述
根据n皇后问题，现在返回n皇后不同的解决方案的数量而不是具体的放置布局。
样例1
输入: n=1
输出: 1
解释:
1:
1
样例2
输入: n=4
输出: 2
解释:
1:
0 0 1 0
1 0 0 0
0 0 0 1
0 1 0 0
2:
0 1 0 0
0 0 0 1
1 0 0 0
0 0 1 0
java题解
dfs暴力搜索即可
public class Solution {
public static int sum;
public int totalNQueens(int n) {
sum = 0;
int[] usedColumns = new int[n];
placeQueen(usedColumns, 0);
return sum;
}
public void placeQueen(int[] usedColumns, int row) {
int n = usedColumns.length;
if (row == n) {
sum ++;
return;
}
for (int i = 0; i < n; i++) {
if (isValid(usedColumns, row, i)) {
usedColumns[row] = i;
placeQueen(usedColumns, row + 1);
}
}
}
public boolean isValid(int[] usedColumns, int row, int col) {
for (int i = 0; i < row; i++) {
if (usedColumns[i] == col) {
return false;
}
if ((row - i) == Math.abs(col-usedColumns[i])) {
return false;
}
}
return true;
}
}
C++题解
dfs暴力搜索即可
class Solution {
public:
int sum;
bool canPut(int row, int col, vector<int> &cols) {
for (int i = 0; i < row; i++) {
if (cols[i] - i == col - row) {
return false;
}
if (cols[i] + i == col + row) {
return false;
}
if (cols[i] == col) {
return false;
}
}
return true;
}
void dfs(int n, int k, vector<int> &cols) {
if (k == n) {
sum++;
return;
}
for (int i = 0; i < n; i++) {
if (!canPut(k, i, cols)) {
continue;
}
cols[k] = i;
dfs(n, k + 1, cols);
}
}
int totalNQueens(int n) {
vector<int> cols(n);
sum = 0;
dfs(n, 0, cols);
return sum;
}
};
python题解
dfs暴力搜索即可
class Solution:
total = 0
n = 0
def attack(self, row, col):
for c, r in self.cols.iteritems():
if c - r == col - row or c + r == col + row:
return True
return False
def search(self, row):
if row == self.n:
self.total += 1
return
for col in range(self.n):
if col in self.cols:
continue
if self.attack(row, col):
continue
self.cols[col] = row
self.search(row + 1)
del self.cols[col]
def totalNQueens(self, n):
self.n = n
self.cols = {}
self.search(0)
return self.total