LeetCode 52. N-Queens II 题解(C++)
2016-10-27 22:22
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LeetCode 52. N-Queens II 题解(C++)
题目描述
Follow up for N-Queens problem.Now, instead outputting board configurations, return the total number of distinct solutions.
![](http://www.leetcode.com/wp-content/uploads/2012/03/8-queens.png)
思路
典型的N皇后问题,使用回溯法。建立一个int数组vec,vec[i] = j的含义为位置(i,j)已经被占领,数组每个元素初始化为0;我们从第一行开始占领(之后第二行占领,第三行占领…),直到行数row等于n,即已经占领完最后一行,则total自加1并return;
每一行占领前,先假设该行可以占领该位置(即vec[row]=column,表示假设位置(row,column)可以占领),之后进行判断,若该位置的列或者对角线已被占领,则放弃该假设的位置,若该位置判别后可以占领,则占领该位置并对下一行继续进行上述的操作。
代码
class Solution { public: bool canTaked(int n, vector<int> &vec, int row) { for (int i = 0; i < row; ++i) { if (vec[i] == vec[row] || abs(vec[i] - vec[row]) == abs(i - row)) { return false; } } return true; } void NQueens(int n, int &total, vector<int> &vec, int row) { if (row == n) { ++total; return; } for (int column = 0; column < n; ++column) { vec[row] = column; if (!canTaked(n, vec, row)) { vec[row] = 0; continue; } NQueens(n, total, vec, row + 1); vec[row] = 0; } } int totalNQueens(int n) { vector<int> vec(n); int total = 0; NQueens(n, total, vec, 0); return total; } };
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