【LeetCode】N-Queens II

题目描述：

Follow up for N-Queens problem.

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

本来我觉得会有dp解法，想了很久没想出来，就拿N-Queens改了下，然后就AC了……都是泪……

class Solution {public:int count;int N;int totalNQueens(int n) {N = n;count = 0;vector<int> rec;solve(n, rec);return count;}void solve(int n, vector<int> &rec){if (!n){count++;return;}int i = rec.size();for (int j = 0; j < N; j++){int m(0);for (m = 0; m < rec.size(); m++)if (rec[m] == j || m + rec[m] == i + j || m - rec[m] == i - j)break;if (m < rec.size())continue;rec.push_back(j);solve(n - 1, rec);rec.pop_back();}}};

还有个在讨论区看到的很神奇的解法记录下，后面再研究……

I found this answer on the Internet. And I spent a while to figure how it works, I put the main idea on the comment. Here it is!

And a picture to illustrate how the bits projection works. 

class Solution {    public:        /* backtrace program using bit-wise operation to speed up calculation.         * 'limit' is all '1's.         * 'h'  is the bits all the queens vertically projected on a row. If h==limit, then it's done, answer++.         * 'r'   is the bits all the queens anti-diagonally projected on a row.         * 'l'   is the bits all the queens diagonally projected on a row.         * h|r|l  is all the occupied bits. Then pos = limit & (~(h|r|l)) is all the free positions.         * p = pos & (-pos)  gives the right most '1'. pos -= p means we will place a queen on this bit          *                             represented by p.         * 'h+p'  means one more queue vertically projected on next row.         * '(r+p)<<1'  means one more queue anti-diagonally projected on next row. Because we are         *                   moving to next row and the projection is skew from right to left, we have to          *                   shift left one position after moved to next row.         * '(l+p)>>1'  means one more queue diagonally projected on next row. Because we are          *                  moving to next row and the projection is skew from left to right, we have to          *                  shift right one position after moved to next row.         */        int totalNQueens(int n) {            ans = 0;            limit = (1<<n) - 1;            dfs(0, 0, 0);            return ans;        }        void dfs(int h, int r, int l) {            if (h == limit) {                ans++;                return;            }            int pos = limit & (~(h|r|l));            while (pos) {                int p = pos & (-pos);                pos -= p;                dfs(h+p, (r+p)<<1, (l+p)>>1);            }        }        int ans, limit;};