题目如下:
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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分析如下:
有了前面的N-QueensII, 这道题目就比较好写了。改变一下输出结果就可以了。
我的代码;
//14ms class Solution { public: void swap(int* array, int a ,int b) { int tmp = array[a]; array[a] = array[b]; array[b] = tmp; } bool collide(int k, int* array) { for (int i = 0; i <k; ++i) { //检查array[k] 和 array[k]之前的数有无冲突 if ((array[k] - k == array[i] - i) || (array[k] + k == array[i] + i)) return true; } return false; } void GenQueens(int k, int n, vector<vector<string> > & result_vec, int* array) { if (k == n){ vector<string> each_vec; for (int j = 0; j < n; ++j) { string s = ""; for (int h = 0; h < array[j]; ++h) { s = s + "."; } s = s + "Q"; for (int h = array[j] + 1; h < n; ++h) { s = s + "."; } each_vec.push_back(s); } result_vec.push_back(each_vec); return; } else { for (int i = k; i < n; ++i) { swap(array, i, k); //如果交换后的当前值引起了冲突,就不进行下一步的dfs if(!collide(k, array)) { GenQueens(k+1, n, result_vec, array); } swap(array, k, i); } } } vector<vector<string> > solveNQueens(int n) { int* array = new int[n]; vector<vector<string> > result_vec; for (int i = 0; i < n; ++i) { array[i] = i; } GenQueens(0, n, result_vec, array); delete [] array; return result_vec; } };