Add solution for N-Queen problem which uses bit-operation

Still uses DFS, optimized for state compression.

Signed-off-by: Hanlin Shi <shihanlin9@gmail.com>
This commit is contained in:
Hanlin Shi
2021-05-31 00:43:43 -07:00
parent ad3dfff270
commit cff768e63c

View File

@ -48,38 +48,48 @@ func generateBoard(n int, row *[]int) []string {
} }
// 解法二 二进制操作法 // 解法二 二进制操作法
// class Solution func solveNQueens2(n int) (res [][]string) {
// { placements := make([]string, n)
// int n; for i := range placements {
// string getNq(int p) buf := make([]byte, n)
// { for j := range placements {
// string s(n, '.'); if i == j {
// s[p] = 'Q'; buf[j] = 'Q'
// return s; } else {
// } buf[j] = '.'
// void nQueens(int p, int l, int m, int r, vector<vector<string>> &res) }
// { }
// static vector<string> ans; placements[i] = string(buf)
// if (p >= n) }
// {
// res.push_back(ans); var construct func(prev []int)
// return ; construct = func(prev []int) {
// } if len(prev) == n {
// int mask = l | m | r; plan := make([]string, n)
// for (int i = 0, b = 1; i < n; ++ i, b <<= 1) for i := 0; i < n; i++ {
// if (!(mask & b)) plan[i] = placements[prev[i]]
// { }
// ans.push_back(getNq(i)); res = append(res, plan)
// nQueens(p + 1, (l | b) >> 1, m | b, (r | b) << 1, res); return
// ans.pop_back(); }
// }
// } occupied := 0
// public: for i := range prev {
// vector<vector<string> > solveNQueens(int n) dist := len(prev) - i
// { bit := 1 << prev[i]
// this->n = n; occupied |= bit | bit<<dist | bit>>dist
// vector<vector<string>> res; }
// nQueens(0, 0, 0, 0, res);
// return res; prev = append(prev, -1)
// } for i := 0; i < n; i++ {
// }; if (occupied>>i)&1 != 0 {
continue
}
prev[len(prev)-1] = i
construct(prev)
}
}
construct(make([]int, 0, n))
return
}