[LeetCode]N-Queens II
2014-07-16 18:11
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题目:给定一个整数n,表示棋盘的规模是n*n,在棋盘上放置n个皇后,要求同一行,同一列,同一条对角线不能有相同的皇后,求出可行的方案数
算法:深度优先搜索 + 剪枝优化
剪枝优化方案如下(拿4皇后为例):
不难发现存在下列规律:
同一条左对角线:row - col + n = 恒定值
同一条右对角线:row + col = 恒定值
算法:深度优先搜索 + 剪枝优化
剪枝优化方案如下(拿4皇后为例):
不难发现存在下列规律:
同一条左对角线:row - col + n = 恒定值
同一条右对角线:row + col = 恒定值
public class Solution { final int CHECK_NUMBER = 3; // check tree points: col, left diagonal and right diagonal final int MAX_NUMBER = 1024; int nSolutions = 0; boolean[][] isVisited = new boolean[CHECK_NUMBER][MAX_NUMBER]; public int totalNQueens(int n) { for (int i=0; i<CHECK_NUMBER; ++i) { isVisited[i] = new boolean[MAX_NUMBER]; } dfs(0, n); return nSolutions; } /** * boolean[0][MAXN]: * * 0: check the col * 1: check the left diagonal * 2. check the right diagonal * */ public void dfs(int row, int nGirds) { if (row == nGirds) { ++nSolutions; return ; } for (int col=0; col<nGirds; ++col) { if (!isVisited[0][col] && !isVisited[1][row-col+nGirds] && !isVisited[2][row+col]) { isVisited[0][col] = true; isVisited[1][row-col+nGirds] = true; isVisited[2][row+col] = true; dfs(row+1, nGirds); isVisited[0][col] = false; isVisited[1][row-col+nGirds] = false; isVisited[2][row+col] = false; } } } }
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