题目: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.
思路: 回溯法,首先建立一个备选答案的vector,每次向其中添加一个经过检查可以添加的元素,如果搜索深度达到n则将备选答案存储
代码:
class Solution {
public:
bool isOK(vector<int> oneSolution, int valToAdd,int k){
for(int j = 0; j < k; j++){
if(oneSolution[j] == valToAdd)
return false;
else if(abs(valToAdd - oneSolution[j]) == abs(j - k))
return false;
}
return true;
}
void backtracking(vector<int> oneSolution, int k, int n, vector<vector<string> > &result){
if (k == n){
vector<string> oneSolutionInChar(n, "");
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) {
if (j == oneSolution[i]) oneSolutionInChar[i] += 'Q';
else oneSolutionInChar[i] += '.';
}
result.push_back(oneSolutionInChar);
}
else{
k = k+1;
for (int i = 0; i<n; i++){
if (!isOK(oneSolution, i, k-1)) continue;
oneSolution[k-1] = i;
backtracking(oneSolution, k, n, result);
}
}
}
vector<vector<string> > solveNQueens(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<string> > result;
if (n == 0) return result;
vector<int> oneSolution(n);
int k = 0;
backtracking(oneSolution, k, n, result);
return result;
}
};