LeetCode N-Queens II

Follow up for N-Queens problem.

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

思路：时间复杂度要求高了，还是每行每行的放，再开一个三维的标记数组，y-x代表了主对角线，y+x代表了负对角线。

class Solution {
public:
    int ans;
    int vis[3][1000];

    void dfs(int cur, int tar) {
        if (cur == tar) {
            ans++;
            return;
        }

        for (int i = 0; i < tar; i++) {
            if (!vis[0][i] && !vis[1][cur+i] && !vis[2][cur-i+tar]) {
                vis[0][i] = vis[1][cur+i] = vis[2][cur-i+tar] = 1;
                dfs(cur+1, tar);
                vis[0][i] = vis[1][cur+i] = vis[2][cur-i+tar] = 0;
            }
        }
    }

    int totalNQueens(int n) {
        ans = 0;
        memset(vis, 0, sizeof(vis));
        dfs(0, n); 
        return ans;
    }
};

LeetCode N-Queens II

原文：http://blog.csdn.net/u011345136/article/details/44197079