Problem Description:
Encountered an algorithm problem, find the number that appears most frequently in the array, and determine whether it exceeds half of the array.
solution
I first thought of 3 schemes. The
first one: use hashSet to pass in data, record the number of false, and finally compare with the array length.
The second type: use two pointers to compare whether the data is duplicated, if some, count+1, and compare with the length of the array.
The third type: The third type is an algorithm to be introduced in this article, Moore voting method. The first two can achieve functions, but there is a problem, both in complexity and space.
Moore voting
public static void main(String[] args) {
/**
* 如果数组中多一半的数都是同一个，则称之为主要元素。给定一个整数数组，找到它的主要元素。若没有，返回-1。
*
* 示例 1：
* 输入：[1,2,5,9,5,9,5,5,5]
* 输出：5
*
* 示例 2：
* 输入：[3,2]
* 输出：-1
*
* 示例 3：
* 输入：[2,2,1,1,1,2,2]
* 输出：2
*
*/
int[] nums = {
1,2,5,9,5,9,5,5,5};
int i = 0;
int count1 = 0;
//遍历整个数组，在遍历的过程中找到出现次数最多的数
for (int num : nums) {
if (count1 == 0) {
i = num;
count1++;
} else {
if (i == num) {
count1++;
} else {
count1--;
}
}
}
int count2 = 0;
//先判断是否有众数
if (count1 <= 0) {
System.out.println(i);
} else {
//确认有众数之后就遍历数组 确定众数出现的次数
for (int num : nums) {
if (num == i) {
count2 ++;
}
}
}
//判断出现的次数是否超过数组的一半
if (count2 > nums.length / 2) {
System.out.println(i);
}
}
to sum up
This algorithm can not only find the number of occurrences more than half the length of the array, but also find one-third and so on.