52. N-Queens II

题目：

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

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链接： http://leetcode.com/problems/n-queens-ii/

题解：

刚昨晚NQueens I， 这换个问题又算一题…不过省了我们很多事。只需要设一个global的count来记录找到解的数目。

Time Complexity – O(nⁿ), Space Complexity – O(n)

public class Solution {
    private int count = 0;    public int totalNQueens(int n) {
        if(n <= 0)
            return count;
        int[] queens = new int[n];      // queens must be a permutation of n
        trySolveNQueens(queens, 0);
        return count;
    }    private void trySolveNQueens(int[] queens, int pos) {
        int len = queens.length;
        if(pos == len)
            count++;
        else {
            for(int i = 0; i < len; i++) {
                queens[pos] = i;
                if(isBoardValid(queens, pos))
                    trySolveNQueens(queens, pos + 1);
            }
        }
    }    private boolean isBoardValid(int[] queens, int pos) {
        for(int i = 0; i < pos; i++) {
            if(queens[i] == queens[pos])        // column conflicts
                return false;
            else if(Math.abs(queens[pos] - queens[i]) == Math.abs(i - pos))  // diagonal conflicts
                return false;
        }        return true;
    }
}

二刷：

Java:

上一题的简化版本，只需要求有多少种solution就可以了。以后再了解一下NQueens的fancy做法

Time Complexity – O(nⁿ), Space Complexity – O(n)

public class Solution {
    private int totalQueens = 0;
    public int totalNQueens(int n) {
        if (n <= 0) {
            return 0;
        }
        int[] queens = new int[n];
        trySolveNQueens(queens, 0);
        return totalQueens;
    }    private void trySolveNQueens(int[] queens, int pos) {
        if (pos == queens.length) {
            totalQueens++;
        } else {
            for (int i = 0; i < queens.length; i++) {
                queens[pos] = i;
                if (isBoardValid(queens, pos)) {
                    trySolveNQueens(queens, pos + 1);
                }
            }
        }
    }    private boolean isBoardValid(int[] queens, int pos) {
        for (int i = 0; i < pos; i++) {
            if (queens[i] == queens[pos]) {
                return false;
            }
            if (Math.abs(queens[i] - queens[pos]) == Math.abs(i - pos)) {
                return false;
            }
        }
        return true;
    }
}

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