标签：
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.."] ]
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思路：
将N-Queens II中的解做一个转化即可。
c++ code:
class Solution {
public:
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> res;
vector<int> nums(n);
for(int i=0;i<n;i++)
nums[i]=i;
permute(res, nums,0);
return res;
}
// 自定义函数
void permute(vector<vector<string>> &res, vector<int> &nums, int pos) {
int n = nums.size();
if(pos==n) {
if(check(nums)) {
vector<string> ans;
for(int i=0;i<n;i++) {
string str(n, '.');
str[nums[i]] = 'Q';
ans.push_back(str);
}
res.push_back(ans);
}
return;
}
for(int i=pos;i<n;i++) {
swap(nums[pos], nums[i]);
permute(res, nums, pos+1);
swap(nums[i], nums[pos]);
}
}
bool check(vector<int> &nums) {
int n = nums.size();
for(int i = 0;i < n;i++) {
for(int j = i + 1;j < n;j++)
// 判断是否在主、副对角线上
if(i-j == nums[i]-nums[j] || j-i == nums[i]-nums[j])
return false;
}
return true;
}
};标签：
原文地址：http://blog.csdn.net/itismelzp/article/details/51615060
