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; }}原文：http://www.cnblogs.com/wangsicongde/p/7576890.html