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N-Queens
模拟退火不会写 0.0
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.."]
]
回溯法，刚开始思路，做一个2层的循环，后来发现其实每行之需要记录一个值，也就是queen的位置就行了。
先生成result strings(只包含queen位置的string)， 然后把strings写成n-queue形式。
对于是否valid，写一个isvalid function. 做2个判断，1 同列是否冲突， 2 对角线是否冲突
代码：
1 public class Solution { 2 public List<List<String>> solveNQueens(int n) { 3 List<List<String>> res = new ArrayList<List<String>>(); 4 if (n <= 0) { 5 return res; 6 } 7 List<String> path = new ArrayList<String>(); 8 int[] row = new int[n]; 9 helper (res, row, n, 0); 10 return res; 11 } 12 private void helper(List<List<String>> resultList, 13 int[] row, 14 int n, 15 int index) { 16 if (n == index) { 17 ArrayList<String> singleResult = translateString(row); 18 resultList.add(singleResult); 19 return; 20 } 21 22 for (int i = 0; i < n; i++) { 23 if (isValid(row, index, i)) { 24 row[index] = i; 25 helper (resultList, row, n, index + 1); 26 row[index] = 0; 27 } 28 } 29 30 } 31 32 private ArrayList<String> translateString(int[] row) { 33 ArrayList<String> resultList = new ArrayList<>(); 34 for (int i = 0; i < row.length; i++) { 35 StringBuilder sb = new StringBuilder(); 36 for (int j = 0; j < row.length; j++) { 37 if (j == row[i]) { 38 sb.append(‘Q‘); 39 } 40 else { 41 sb.append(‘.‘); 42 } 43 } 44 resultList.add(sb.toString()); 45 } 46 return resultList; 47 } 48 49 private boolean isValid(int[] row, int rowNum, int columnNum) { 50 for (int i = 0; i < rowNum; i++) { 51 if (row[i] == columnNum) { 52 return false; 53 } 54 //对角线检查 55 if (Math.abs(row[i] - columnNum) == Math.abs(i - rowNum)) { 56 return false; 57 } 58 } 59 return true; 60 } 61 }
--------------------------我是分割线-------------------
N-Queens II
Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
注意建立的res 是静态变量，在主方法里面赋值，防止它是dirty的
1 public class Solution { 2 static int res; 3 public int totalNQueens(int n) { 4 if (n <= 0) { 5 return 0; 6 } 7 int[] rows = new int[n]; 8 int index = 0; 9 res = 0; 10 helper(0, rows); 11 return res; 12 13 } 14 15 private static void helper(int index, int[] rows) { 16 int n = rows.length; 17 if (index == n) { 18 res++; 19 return; 20 } 21 for (int i = 0; i < n; i++) { 22 if (isValid(rows, index, i)) { 23 rows[index] = i; 24 helper(index + 1, rows); 25 } 26 } 27 } 28 29 private static boolean isValid(int rows[], int rowNum, int columnNum) { 30 for (int i = 0; i < rowNum; i++) { 31 if (rows[i] == columnNum) { 32 return false; 33 } 34 if (Math.abs(rows[i] - columnNum) == Math.abs(i - rowNum)) { 35 //columnNum1 - columnNum2 = rowNum1 - rowNum2 平行！ 36 return false; 37 } 38 } 39 return true; 40 } 41 }
回溯法(backtracking) 题目整理--------part2
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原文地址：http://www.cnblogs.com/jiangchen/p/5931754.html
