52.多皇后问题第二弹

N-Queens II

问题描述：

Follow up for N-Queens problem.

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

测试代码（C++）：

int totalNQueens(int n) {    vector<bool> col(n, true);    vector<bool> anti(2*n-1, true);    vector<bool> main(2*n-1, true);    vector<int> row(n, 0);    int count = 0;    dfs(0, row, col, main, anti, count);    return count;}void dfs(int i, vector<int> &row, vector<bool> &col, vector<bool>& main, vector<bool> &anti, int &count) {        if (i == row.size()) {            count++;            return;        }       for (int j = 0; j < col.size(); j++) {         if (col[j] && main[i+j] && anti[i+col.size()-1-j]) {             row[i] = j;              col[j] = main[i+j] = anti[i+col.size()-1-j] = false;             dfs(i+1, row, col, main, anti, count);             col[j] = main[i+j] = anti[i+col.size()-1-j] = true;      }    }}

测试代码（python）：

class Solution(object):    def totalNQueens(self, n):        self.res = 0        self.dfs([-1]*n, 0)        return self.res    def dfs(self, nums, index):        if index == len(nums):            self.res += 1            return         for i in xrange(len(nums)):            nums[index] = i            if self.valid(nums, index):                self.dfs(nums, index+1)    def valid(self, nums, n):        for i in xrange(n):            if nums[i] == nums[n] or abs(nums[n]-nums[i]) == n-i:                return False        return True

性能：