众数问题

众数问题算法有很多实现，众数问题的实现选择要考虑多个方面，这里用到了快速排序算法思想，将每次递归计数到返回值Map里，再与之比较

public class ModeAlgorithmT {private int[] arrTemp;public ModeAlgorithmT(int[] arr) {// TODO Auto-generated constructor stubthis.arrTemp = arr;Map<Integer, Integer> xx = quickSort(arrTemp, 0, arr.length - 1);Iterator it = xx.keySet().iterator();int key, count = 0, mode = 0;while (it.hasNext()) {key = Integer.parseInt(it.next().toString());if (xx.get(key) > count) {mode = key;count = xx.get(key);}System.out.println(xx.size());}System.out.println("众数是" + mode);System.out.println("重数是" + count);}public Map<Integer, Integer> quickSort(int[] arr, int lowIndex,int highIndex) {int low = lowIndex;int high = highIndex;Map<Integer, Integer> xx = new HashMap<Integer, Integer>();int mid = 0, count = 0;if (highIndex > lowIndex) {mid = arr[lowIndex + (highIndex - lowIndex) / 2];while (low <= high) {while ((low < highIndex) && (arr[low] < mid)) {++low;}while ((high > lowIndex) && (arr[high] > mid)) {--high;}if (low <= high) {int temp = arr[low];arr[low] = arr[high];arr[high] = temp;++low;--high;}if (high != low) {//避免重复计数

if (arr[high] == mid) {count++;}if (arr[low] == mid) {count++;}} else {if (arr[high] == mid) {count++;}}}if (lowIndex < high) {Map<Integer, Integer> temp = quickSort(arr, lowIndex, high);int key = temp.keySet().iterator().next();if (temp.get(key) > count) {count = temp.get(key);mid = key;}}if (low < highIndex) {Map<Integer, Integer> temp = quickSort(arr, low, highIndex);int key = temp.keySet().iterator().next();if (temp.get(key) > count) {//比较递归调用返回的重数与众数count = temp.get(key);mid = key;}}}xx.put(mid, count);//存入当前众数与重数return xx;}}