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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.
解题思路：经典的八皇后问题，由于之前在《JAVA语言程序设计》中exerice6-22中做过这种问题，代码直接拿来修改下即可，JAVA实现如下：
static public List<String[]> solveNQueens(int n) {
List<String[]> list=new ArrayList<String[]>();
if(n==1){
String[] str={"Q"};
list.add(str);
return list;
}
// int count = 0;
int[] queens = new int[n]; // queens are placed at (i, queens[i])
for (int i = 0; i < n; i++)
queens[i] = -1;
queens[0] = 0;
int k = 1;
while (k >=0) {
int j = findPosition(k, queens,n);
if (j ==-1) {
queens[k] = -1;
k--; // back track to the previous row
} else {
queens[k] = j;
if (k == n-1) {
// count++;
String[] queenArray=new String[n];
for(int i=0;i<n;i++){
StringBuilder sb=new StringBuilder();
for(int j2=0;j2<n;j2++){
if(j2==queens[i])
sb.append(‘Q‘);
else sb.append(‘.‘);
}
queenArray[i]=new String(sb);
}
list.add(queenArray);
}
else {
k++;
}
}
}
return list;
}
public static int findPosition(int k, int[] queens,int n) {
int start = queens[k] == -1 ? 0 : queens[k] + 1;
for (int j = start; j < n; j++) {
if (isValid(k, j, queens,n))
return j;
}
return -1;
}
public static boolean isValid(int k, int j, int queens[],int n) {
for (int i = 0; i < k; i++)
if (queens[i] == j)
return false;
for (int row = k - 1, column = j - 1; row >= 0 && column >= 0; row--, column--)
if (queens[row] == column)
return false;
for (int row = k - 1, column = j + 1; row >= 0 && column <= n-1; row--, column++)
if (queens[row] == column)
return false;
return true;
}
Java for LeetCode 051 N-Queens
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原文地址：http://www.cnblogs.com/tonyluis/p/4505145.html
