### Question

### Solution

Given an array of size *n*, find the majority element. The majority element is the element that appears more than `? n/2 ?`

times.

You may assume that the array is non-empty and the majority element always exist in the array.

**Credits:**

Special thanks to @ts for adding this problem and creating all test cases.

### Analysis

I have already covered this question in [LintCode] Majority Number. However, I have also discovered **a few other very interesting solution**. Check below.

### Solution

The best solution is of course the voting algorithm. Read code below.

Second solution presented by offical answer, is to do sorting:

```
public int majorityElement(int[] num) {
if (num.length == 1) {
return num[0];
}
Arrays.sort(num);
return num[num.length / 2];
}
```

Third interesting solution is doing bit manipulation. Read Yanyulin’s blog for more.

### Code

```
public class Solution {
public int majorityElement(int[] num) {
if (num == null || num.length == 0) {
return 0;
}
int major = num[0];
long count = 1;
for (int i = 1; i < num.length; i++) {
if (major == num[i]) {
count++;
} else if (count == 0) {
major = num[i];
count = 1;
} else {
count--;
}
}
return major;
}
}
```