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 = 1
                ans = nums[i]
            elif nums[i] == ans:
                cnt += 1
            else:
                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 = 0
        i = 1
        while i<len(nums):
            if nums[i] == nums[i-1]: 
                if nums[i] == ans: # check if is prev ans
                    dup_flag += 2                
                else:
                    if dup_flag >= 2:
                        dup_flag -= 2
                    else:
                        dup_flag = 2
                        ans = nums[i]                                           
            i += 2
        if 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;
    } 
};