笔试算法题目，奶牛排队喝水

Posted 2020-10-08 王思聪6

import java.util.*;

/**

 * Created by Daxin on 2017/8/19.

 * <p/>

 * 奶牛排队饮水问题

 * 输入：n牛的数目，然后n个整数表示牛的序号

 * 输出：输出交换最少次数

 * <p/>

 * 例如一个测试用例：9<br>

 * 2,2,1,3,3,3,2,3,1<br>

 * 输出：4

 *

 *在线题目地址：http://www.hustoj.com/oj/problem.php?id=1056

 *

 *

 */

public class CattleSortWater {

    public static void main(String[] args) {

//        int[] nums = {2, 2, 1, 3, 3, 3, 2, 3, 1};

        Scanner cin = new Scanner(System.in);

        int n = cin.nextInt();

        int[] nums = new int[n];

        for (int i = 0; i < n; i++) {

            nums[i] = cin.nextInt();

        }

        System.out.println(solve(nums));

//        System.out.println(Arrays.toString(nums));

    }

    public static int solve(int[] nums) {

        int result = 0;

        Map<Integer, Integer> wcount = new TreeMap<>();

        Map<Integer, Integer> range = new TreeMap<>();

        for (int num : nums) {

            Integer tmp = wcount.get(num);

            wcount.put(num, tmp == null ? 1 : tmp + 1);

        }

        Set<Integer> set = wcount.keySet();

        int pre = 0;

        for (int i : set) {

            int tmp = wcount.get(i) + pre;

            range.put(i, tmp);

            pre = tmp;

        }

        for (int num : set) {

            int times = wcount.get(num);

            int ran = range.get(num);

            for (int i = ran - 1, t = times; t > 0; t--, i--) {

                if (nums[i] != num) {

                    int r = get(nums, range.get(nums[i]), wcount.get(nums[i]), num);

                    if (r != -1) { //在nums[i] 区间中寻找到了num

                        int tmp = nums[r];

                        nums[r] = nums[i];

                        nums[i] = tmp;

                        result++;

                    } else {// 在希望区间没有找到，进行余下遇见全扫描

                        int rr = getRange(nums, ran, num,range.get(nums[i])-wcount.get(nums[i]),range.get(nums[i]));

                        if (rr != -1) {

                            int tmp = nums[rr];

                            nums[rr] = nums[i];

                            nums[i] = tmp;

                            result++;

                        }

                    }

                }

            }

        }

        return result;

    }

    /**

     *

     * @param nums 数组

     * @param range 结束的下标

     * @param times 出现的次数，range-times就是起始下标

     * @param expect 希望查找的希望值

     * @return

     */

    public static int get(int[] nums, int range, int times, int expect) {

        for (int i = range - 1, t = times; t > 0; t--, i--) {

            if (nums[i] == expect) {

                return i;

            }

        }

        return -1;

    }

    /**

     *

     * @param nums 数组

     * @param start 查找的起始位置，结束位置是数组的结束

     * @param expect 期望寻找到的值

     * @param exceptStart 排除区域的起始下标

     * @param exceptEnd 排除区域的结束下标

     * @return 返回下标

     */

    public static int getRange(int[] nums, int start, int expect,int exceptStart,int exceptEnd) {

        for (; start < nums.length; start++) {

            if(start>=exceptStart&&start<exceptEnd)

                continue;

            if (nums[start] == expect) {

                return start;

            }

        }

        return -1;

    }

}