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.
分析:先将数组排序,因为majority element出现次数超过⌊ n/2 ⌋,即超过数组长度的一半。
代码:
public class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums);//时间复杂度O(nlogn),代码虽然简洁,但不是最优解 return nums[nums.length / 2]; } }
本算法的基本思想是:遍历数组,从起始元素开始,如果下一元素与上一元素相同,计数(count++)加一,如果与上一元素不同,则减一,直到数组只剩下一种元素。
代码:
public class Solution { public int majorityElement(int[] nums) { // Arrays.sort(nums);//时间复杂度O(nlogn),代码虽然简洁,但不是最优解 // return nums[nums.length / 2]; int count = 0; int res = nums[0]; for(int i = 1; i < nums.length; i++){ if(nums[i] == res){ count++; }else{ if(count > 0){ count--; }else{ res = nums[i]; } } } return res; } }