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.
思路:先将这个数组排序,然后统计出现次数最多的那个数,时间复杂度为排序的o(nlgn)+遍历的o(n),所以时间复杂度是o(nlgn),额外的空间为0
代码如下(已通过leetcode)
public class Solution {
public int majorityElement(int[] nums) {
if(nums.length>Integer.MAX_VALUE) return 0;
Arrays.sort(nums);
int majele=nums[0];
int max=1;
int i=0;
int j=1;
while(i<nums.length) {
int count=1;
for(j=i+1;j<nums.length;j++) {
if(nums[i]==nums[j]) count++;
else break;
}
if(count>max) {
max=count;
majele=nums[i];
}
i=j;
}
return majele;
}
}