[LeetCode]52. N-Queens II

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Problem

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 the number of distinct solutions to the n-queens puzzle. 

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Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

Solution

Analysis

这道题的解法和51题相同,都是使用回溯法. 只不过这里,不需要记录所有可能的解,仅仅需要返回解的个数即可. 偷懒的话,可以直接使用上一题的解法,最后返回可能解的个数,也就是数组的容量大小.

这里,只需要返回可能解的个数, 所以不需要记录所有可能的解,仅仅需要一个可变的统计量count, 同时一个数组记录之前皇后的摆放位置, 这里使用pair结构来记录每个皇后的坐标(二维[x, y]). 具体代码如下:

Code

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class Solution {
public:
    int totalNQueens(int n) {
        vector<pair<int, int>> points;
        int count = 0;
        
        dfs(n, 0, points, count);
        
        return count;
    }
private:
    bool isAttack(const pair<int, int> &p1, const pair<int, int> &p2){
        if (p1.second == p2.second || abs(p1.first - p2.first) == abs(p1.second - p2.second))
            return true;
        return false;
    }
    void dfs(int n, int i, vector<pair<int, int>>& points, int &count){
        if (i == n){
            ++count;
            return;
        }
        
        for (int j=0; j< n; ++j){
            pair<int, int> candidate = make_pair(i, j);
            bool isValid = true;
            
            for(auto p: points){
                if (isAttack(p, candidate)){
                    isValid = false;
                    break;
                }
            }
            
            if (isValid){
                points.push_back(candidate);
                
                dfs(n, i+1, points, count);
                
                points.pop_back();// 回溯
            }
        }
    }
};
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