topic description
There is a number in the array that occurs more than half of the length of the array, please find this number. For example, input an array {1,2,3,2,2,2,5,4,2} with a length of 9. Since the number 2 appears 5 times in the array, which exceeds half the length of the array, 2 is output. Output 0 if not present.
train of thought
1. Practice the use of HashMap, and the statistics are over;
2. Sort, take the median;
3. Invincible-Moore voting method.
Moore voting method:
Sum of votes: Since the number of times the mode appears exceeds half of the length of the array; if the number of votes for the mode is +1 and the number of votes for the non-moderate is −1, then there must be a sum of votes of all numbers > 0.
Positive and negative offset of votes: Let the mode of the array nums be x and the length of the array be n. If the sum of votes of the first a numbers of nums = 0, then the sum of votes of the last (na) numbers in the array must still be >0 (that is, the mode of the last (na) numbers is still x).
It's a very clear idea, and it's true.
the code
1. HashMap practice.
import java.util.HashMap;
import java.util.Map;
class Solution {
public static int majorityElement(int[] nums) {
// 创建hashmap, 数-次数
HashMap<Integer, Integer> hash = counter(nums);
// 遍历，选出长度大于一半的
int res = MapEntry(hash, nums);
return res;
}
public static HashMap<Integer, Integer> counter(int[] nums) {
HashMap<Integer, Integer> count = new HashMap<Integer, Integer>();
for (int num : nums) {
if (count.containsKey(num)) {
int old_v = count.get(num);
count.put(num, old_v + 1);
} else {
count.put(num, 1);
}
}
return count;
}
public static int MapEntry(HashMap<Integer, Integer> hashmap, int[] nums) {
int res = 0;
for (Map.Entry<Integer, Integer> entry : hashmap.entrySet()) {
if (entry.getValue() > (nums.length / 2)) {
res = entry.getKey();
}
}
return res;
}
}
2. Don't write the sorting;
3. Moore voting method
class Solution {
public int majorityElement(int[] nums) {
int x = 0, votes = 0;
for(int num : nums){
if(votes == 0) x = num;
votes += num == x ? 1 : -1;
}
return x;
}
}