【LeetCode051】N皇后问题，回溯法，加上一个栈维护结果

同【LeetCode052】

只是要具体的结果，那么只需多用一个全局的栈来存结果，每次找到一个，添加到结果队列ansList里即可

注意，vector<vector<string> > ansList;

>>之间要有一个空格，不然两个>会被认为是>>，会报错

AC代码

#include<stdio.h>
#include<string>
#include<vector>
#include<iostream>
using namespace std;

int col[101];
int x1[202];
int x2[202];
int ans = 0;
int ansStack[101];
int top = 0;
vector<vector<string> > ansList;

class Solution {
public:
vector<vector<string> > solveNQueens(int n) {
ans = 0;
top = 0;
ansList.clear();
memset(col, 0, sizeof(col));
memset(x1, 0, sizeof(x1));
memset(x2, 0, sizeof(x2));

solve(0, n);
// printf("%d\n", ans);
return ansList;
}
void solve(int ri, int n) {
if (ri == n) {
vector<string> oneAns;
for (int i = 0 ; i < n; i++) {
string str = "";
for (int j = 0; j < n; j++) {
if (j == ansStack[i]) {
str += "Q";
} else {
str += ".";
}
}
oneAns.push_back(str);
}
ansList.push_back(oneAns);
ans++;
return;
}

int ci;
for (ci = 0; ci < n; ci++) {
if (col[ci] || x1[ri + ci] || x2[ri + n - 1 - ci]) continue;

col[ci] = 1;
x1[ri + ci] = 1;
x2[ri + n - 1 - ci] = 1;
ansStack[top++] = ci;
solve(ri + 1, n);
top--;
col[ci] = 0;
x1[ri + ci] = 0;
x2[ri + n - 1 - ci] = 0;
}
}
};

int main() {
Solution s;
ansList = s.solveNQueens(4);
for (int i = 0; i < ansList.size(); i++) {
for (int j = 0; j < ansList[i].size(); j++){
cout << ansList[i][j] << endl;
}
}
return 0;
}