算法笔记_131:出现次数超过一半的数(Java)

1 问题描述

2 解决方案

2.1 每次删除两个不同的数

package com.liuzhen.practice;

public class Main {

public int getResult(int[] A) {
//找到两个不相等的元素，将这两个元素变为0
for(int start = 0, i = 1;i < A.length;i++) {
while(A[start] == 0) {
start++;
}
if(A[start] != A[i] && start < i) {
A[start] = 0;
A[i] = 0;
start++;
}
}
int result = 0;
for(int i = 0;i < A.length;i++) {
if(A[i] != 0) {
result = A[i];
break;
}
}
return result;
}

public static void main(String[] args) {
Main test = new Main();
int[] A = {1,2,3,4,5,6,7,1,1,2,3,2,2,2,2,2};
System.out.println(test.getResult(A));
}
}

2

2.2 记录两个值

package com.liuzhen.practice;

public class Main1 {

public int FindOneNumber(int[] A) {
int result = A[0];
int count = 1;
for(int i = 1;i < A.length;i++) {
if(count == 0) {
result = A[i];
count = 1;
} else {
if(A[i] == result)
count++;
else
count--;
}
}
return result;
}

public static void main(String[] args) {
Main1 test = new Main1();
int[] A = {1,2,3,4,5,6,7,1,1,2,3,2,2,2,2,2};
System.out.println(test.FindOneNumber(A));
}
}

2