leetcode 169. Majority Element

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.

两种解法，一种是排序后取中位数！

另外一种是用计数的方式，Moore voting algorithm 据说是叫这个，和选票一半以上的人当选一样，计票方法：如果是一张反对票一张赞成票则互相抵消。

class Solution(object):def majorityElement(self, nums):""":type nums: List[int]:rtype: int"""ans = nums[0]cnt = 1        for i in xrange(1, len(nums)):if cnt == 0:cnt = 1ans = nums[i]elif nums[i] == ans:cnt += 1else:cnt -= 1        return ans

另外我自己的解法是每两位计数一次，计数重复出现的数，见注释示例：

class Solution(object):def majorityElement(self, nums):""":type nums: List[int]:rtype: int"""#nums.sort()#return nums[len(nums)>>1]# 1 1 2=>1==1 ans=1, dup=2, last 2 dump it# 2 1 1=>2!=1, dup=0, ans=1# 1 1 1 2=>1=1, dup=2, ans=1, 1!=2 # 1 2 3 1 1 1=>1!=2, dup=0, 3!=1, dup=0, 1==1, dup=2, ans=1# 1 1 3 3 2 1 1=>1==1,dup=2, 3==3, dup-=2,dup=0, 2!=0, 1 is answer# 1 1 1 1 3 3 3 3 2 1 1=>ans = nums[0]dup_flag = 0i = 1while i<len(nums):if nums[i] == nums[i-1]: if nums[i] == ans: # check if is prev ansdup_flag += 2                else:if dup_flag >= 2:dup_flag -= 2else:dup_flag = 2ans = nums[i]                                           i += 2if len(nums) & 1 and dup_flag == 0:return nums[-1]return ans

最后一种是位运算：

Bit Manipulation

Another nice idea! The key lies in how to count the number of 1's on a specific bit. Specifically, you need a mask with a 1 on the i-the bit and 0 otherwise to get the i-th bit of each element in nums. The code is as follows.

class Solution {
public:int majorityElement(vector<int>& nums) { int major = 0, n = nums.size(); for (int i = 0, mask = 1; i < 32; i++, mask <<= 1) { int bitCounts = 0; for (int j = 0; j < n; j++) { if (nums[j] & mask) bitCounts++; if (bitCounts > n / 2) { major |= mask; break; } } } return major; } };

转载于:https://www.cnblogs.com/bonelee/p/8654829.html

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