Backtracking

回溯：当把问题分成若干步骤并递归求解时，如果当前步骤没有合法选择，则函数将返回上一级递归调用，这种现象称为回溯。

如果在回溯法中使用了辅助的全局变量，则一定要及时把它们恢复原状。特别地，若函数有多个出口，则需在每个出口处回复被修改的值。

Leetcode 51. N-Queens

给定$n$，输出$n$皇后问题的所有解。

class Solution {
public:
    vectorstring>> solveNQueens(int n) {
        vectorint>> vis(3, vector<int>(2 * n, 0));
        vector<int> C(n, 0);
        vectorstring>> res;
        search(vis, 0, n, res, C);
        return res;
    }
    
    void search(vectorint>>& vis, int cur, int n, vectorstring>>& res, vector<int>& C){
        if(cur == n){
            vector<string> tmp(n, string(n, '.'));
            for(int i = 0; i < n; ++i){
                tmp[i][C[i]] = 'Q';
            }
            res.push_back(tmp);
        }
        else{
            for(int i = 0; i < n; ++i){
                if(!vis[0][i] && !vis[1][cur + i] && !vis[2][cur - i + n]){
                    C[cur] = i;
                    vis[0][i] = vis[1][cur + i] = vis[2][cur - i + n] = 1;
                    search(vis, cur + 1, n, res, C);
                    vis[0][i] = vis[1][cur + i] = vis[2][cur - i + n] = 0;
                }
            }
        }
    }
};

Leetcode 52. N-Queens II

给定$n$，输出$n$皇后解的个数。

class Solution {
public:
    int totalNQueens(int n) {
        vectorint>> vis(3, vector<int>(2 * n, 0));
        int total = 0;
        search(vis, total, 0, n);
        return total;
    }
    
    void search(vectorint>>& vis, int& total, int cur, int n){
        if(cur == n) total++;
        else{
            for(int i = 0; i < n; ++i){
                if(!vis[0][i] && !vis[1][cur + i] && !vis[2][cur - i + n]){
                    vis[0][i] = vis[1][cur + i] = vis[2][cur - i + n] = 1;
                    search(vis, total, cur + 1, n);
                    vis[0][i] = vis[1][cur + i] = vis[2][cur - i + n] = 0;
                }
            }
        }
    }
};