每天刷个算法题20160519：回溯法解八皇后

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http://blog.csdn.net/xiaofei_it/article/details/51502622

为了防止思维僵化，每天刷个算法题。已经刷了几天了，现在发点代码。

我已经建了一个开源项目，每天的题目都在里面：

https://github.com/Xiaofei-it/Algorithms

绝大部分算法都是我自己写的，没有参考网上通用代码。读者可能会觉得有的代码晦涩难懂，因为那是我自己的理解。

最近几天都是在写一些原来的东西，大多数是非递归。以后准备刷点DP、贪心之类的题。

下面是回溯法八皇后：

/**
 *
 * Copyright 2016 Xiaofei
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 * http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 *
 */

package xiaofei.algorithm;

/**
 * Created by Xiaofei on 16/5/19.
 */
public class EightQueensPuzzle {

    /**
     * 八皇后（回溯法）
     * pos[n]
     * 竖排：值相同
     * 左上到右下：i - j = pos[i] - pos[j]  -->  i - pos[i] = j - pos[j]
     * 左下到右上：i - j = pos[j] - pos[i]  -->  i + pos[i] = j + pos[j]
     *
     * 每次都要O(n^2)判断有点蛋疼，不知道有没有优化方案。
     * 现在自己优化一下：
     * hash判断
     * 竖排：hash[SIZE]
     * 左上到右下：i - pos[i] = j - pos[j] 从-(SIZE-1)到(SIZE-1) 为了从0开始，偏移(SIZE-1)
     * 左下到右上：i + pos[i] = j + pos[j] 从0到2*(SIZE-1)
     */

    private static final int SIZE = 8;

    private static int number = 0;

    private static void print(int[] pos) {
        System.out.println(++number);
        int length = pos.length;
        for (int i = 0; i < length; ++i) {
            System.out.print(" " + pos[i]);
        }
        System.out.println();
        for (int i = 0; i < length; ++i) {
            for (int j = 0; j < pos[i]; ++j) {
                System.out.print('+');
            }
            System.out.print('*');
            for (int j = pos[i] + 1; j < SIZE; ++j) {
                System.out.print('+');
            }
            System.out.println();
        }
    }

    public static void calculate() {
        int[] pos = new int[SIZE];
        boolean[] hash1 = new boolean[SIZE];
        boolean[] hash2 = new boolean[SIZE * 2 + 1];
        boolean[] hash3 = new boolean[SIZE * 2 + 1];
        int last = 0;
        pos[last] = -1;
        while (last >= 0) {
            if (last == SIZE) {
                print(pos);
                --last;
            } else {
                boolean flag = false;
                if (pos[last] >= 0) {
                    hash1[pos[last]] = false;
                    hash2[last - pos[last] + SIZE - 1] = false;
                    hash3[last + pos[last]] = false;
                }
                while (pos[last] < SIZE - 1) {
                    ++pos[last];
                    if (!hash1[pos[last]] && !hash2[last - pos[last] + SIZE - 1] && !hash3[last + pos[last]]) {
                        hash1[pos[last]] = true;
                        hash2[last - pos[last] + SIZE - 1] = true;
                        hash3[last + pos[last]] = true;
                        ++last;
                        if (last < SIZE) {
                            pos[last] = -1;
                        }
                        flag = true;
                        break;
                    }
                }
                if (!flag) {
                    --last;
                }
            }
        }
    }
}

每天刷个算法题20160519：回溯法解八皇后

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原文地址：http://blog.csdn.net/xiaofei_it/article/details/51502622