标签：app style this bsp [] log test int obj
这道题为简单题
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.
1、我是利用字典，空间复杂度较高，遍历列表，如果已经该元素已经存在于字典中，那么键值+1，否则创建键值对，并且每次遍历比较最大键值m，并保留该值的对应元素n,最后返回n
2、大神就只用了一个变量count计数，因为有超过一半的数都是该值，所以count==0那儿，到最后count不会小于等于0
我的：
1 class Solution(object): 2 def majorityElement(self, nums): 3 """ 4 :type nums: List[int] 5 :rtype: int 6 """ 7 a = {} 8 m = 0 9 n = 0 10 for i in nums: 11 if i in a: 12 a[i] += 1 13 else: a[i] = 1 14 if a[i] > m: 15 m = a[i] 16 n = i 17 return n
大神：
1 public class Solution { 2 public int majorityElement(int[] num) { 3 4 int major=num[0], count = 1; 5 for(int i=1; i<num.length;i++){ 6 if(count==0){ 7 count++; 8 major=num[i]; 9 }else if(major==num[i]){ 10 count++; 11 }else count--; 12 13 } 14 return major; 15 } 16 }
标签：app style this bsp [] log test int obj
原文地址：http://www.cnblogs.com/liuxinzhi/p/7551153.html
