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;
}
}