169. Majority Element My Submissions Question
2016-02-27 10:49
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Total Accepted: 95925 Total Submissions: 239241 Difficulty: Easy
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.
O(n*n):判断每个元素是否是majority
O(n): 利用HashTable存储每个元素的个数,空间复杂度高
O(nlogn):对数组排序,majority肯定在n/2位置处。
O(nlogn):分而治之,分成两分A和B,如果A和B的majority相等,则其为整个数组的majority;否则,分别扫描计算A,B的majority元素个数(O(2n)),所以时间复杂度T(n) = T(n/2) + 2n = O(nlogn)。
O(n):两两删除数组中两个不同的元素,最后剩下的就是多的元素。代码如下:
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.
O(n*n):判断每个元素是否是majority
O(n): 利用HashTable存储每个元素的个数,空间复杂度高
O(nlogn):对数组排序,majority肯定在n/2位置处。
O(nlogn):分而治之,分成两分A和B,如果A和B的majority相等,则其为整个数组的majority;否则,分别扫描计算A,B的majority元素个数(O(2n)),所以时间复杂度T(n) = T(n/2) + 2n = O(nlogn)。
O(n):两两删除数组中两个不同的元素,最后剩下的就是多的元素。代码如下:
int majorityElement(vector<int>& nums) { if (nums.empty()) return -1; int candidate, count = 0; for (int i = 0; i < nums.size(); ++i) { if (count == 0) { candidate = nums[i]; ++count; } else { if (candidate == nums[i]) ++count; else --count; } } return candidate; }
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