#
一天一道LeetCode系列 ##(一)题目The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space >respectively. For example, There exist two distinct solutions to the 4-queens puzzle: [ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ] ##(二)解题 不玩国际象棋还真不知道这题的规则。百度了好久才明白。 主要有以下三个:
- 同一行上只能有一个皇后
- 同一列上只能有一个皇后
- 两个皇后之间不能处在同一条对角线上 具体解法看代码:
/*
首先利用一个数row和数组a[i]确保每一行每一列只有一个皇后
然后利用一个vector存储已摆放皇后的位置坐标,每摆一个皇后就与已摆放的皇后进行比较,如果在一条对角线上就不能摆
*/
class Solution {
public:
vector<vector<string>> ret;
vector<pair<int, int>> queens;//存放已摆放的皇后的坐标值
vector<vector<string>> solveNQueens(int n) {
int *a = new int[n];//确保每一列只有一个皇后
memset(a,0,n*sizeof(int));
vector<string> res;
backtrc(res, a, 0, n);
return ret;
}
bool isValid(vector<pair<int,int>> queens , int row,int col)//
{
if (queens.empty()) return true;
for (int i = 0; i < queens.size();i++)
{
if (abs(row- queens[i].first) == abs(col-queens[i].second))
{
return false;
}
}
return true;
}
void backtrc(vector<string> res, int a[], int row, int n)//row确保每一行只有一个皇后
{
if (row == n)//如果摆放完n行,则退出
{
ret.push_back(res);
return;
}
for (int i = 0; i < n; i++)
{
if (a[i] == 0&&isValid(queens, row, i))//保证了同一行,同一列,同一对角线只有一个Q
{
a[i] = 1;
string tmp(n, '.');
tmp[i] = 'Q';
res.push_back(tmp);
queens.push_back(pair<int, int>(row, i));
backtrc(res, a, row + 1, n);
//回溯
a[i] = 0;
queens.pop_back();
res.pop_back();
}
}
}
};