LeetCode--N-Queens II
2017-09-01 10:46
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Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
思路:回溯法。由于不用放置整个棋盘上的元素,所以类似数独那一题,只用判断列和两个对角线是否可以放置即可。还是深度优先搜索,逐行逐列的放置,当最后一行也可以放置时计数加1。
Now, instead outputting board configurations, return the total number of distinct solutions.
思路:回溯法。由于不用放置整个棋盘上的元素,所以类似数独那一题,只用判断列和两个对角线是否可以放置即可。还是深度优先搜索,逐行逐列的放置,当最后一行也可以放置时计数加1。
class Solution { public: int totalNQueens(int n) { vector<bool>col(n,true); vector<bool>main(2*n-1,true); vector<bool>anti(2*n-1,true); int count=0; dfs(0,count,n,col,main,anti); return count; } void dfs(int i,int &count,int n,vector<bool>&col,vector<bool>&main,vector<bool>&anti){ if(i==n){ count++; return; } for(int j=0;j<n;j++){ if(col[j]&&main[i+j]&&anti[i+n-1-j]){ col[j]=main[i+j]=anti[i+n-1-j]=false; dfs(i+1,count,n,col,mai 4000 n,anti); col[j]=main[i+j]=anti[i+n-1-j]=true; } } } };
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