编程之美2.3 寻找发帖“水王”

//题目1：找到数组中出现次数超过一半的元素//解法：使用另外不到1/2的元素与所求元素进行抵消//整个数组抵消不同的两个数，则超过1/2的元素还是原来的元素public class Main {        public static void main(String[] args) throws Exception {      System.out.println(findNum(new int[]{1,2,2,3,4,4,4,4,4}));    }        public static int findNum(int[] input) throws Exception {          int num = input[0];        int count = 1;        for(int i = 1;i<input.length;i++){        if(count == 0){        num = input[i];        count = 1;        continue;        }        if(input[i] == num){        count++;        }else{        count--;        }        }        return num;    }    }  //题目2：找到数组中3个出现次数都超过总数1/4的元素//解法：与题目1思路基本相同，建立三个num，让这三个num中的数字与另外不到1/4的数字进行抵消public class Main {  public static void main(String[] args) throws Exception {  System.out.println(findNum(new int[]{1,1,2,2,2,3,3,3,4,4,4}));}  public static int[] findNum(int[] input) throws Exception {  int num1 = -1;int num2 = -1;int num3 = -1;int count1 = 0;int count2 = 0;int count3 = 0;for(int i = 0;i<input.length;i++){if(input[i] == num1 && count1!=0){count1++;continue;}else if(input[i] == num2 && count2!=0){count2++;continue;}else if(input[i] == num3 && count3!=0){count3++;continue;}if(count1 == 0){num1 = input[i];count1 = 1;continue;}if(count2 == 0){num2 = input[i];count2 = 1;continue;}if(count3 == 0){num3 = input[i];count3 = 1;continue;}count1--;count2--;count3--;}return new int[]{num1,num2,num3};}  }