Wins the Offer (28): More than half of the number of array numbers appear

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First, the primer

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Second, the title

There are a number of array number that appears more than half the length of the array, find this number. For example, a length of the input array 9 {1,2,3,2,2,2,5,4,2}. Since the number 2 appears five times, more than half the length of the array in the array, the output 2. If there is 0 output.

1, ideas

1 idea: to sort the array to find the number in the middle of the array. If the number of times a number appears more than half the length of the array, then after ordering this number is certainly among the best, then count the number of ways to view the number of occurrences is greater than half the length of the array, then it returns this number, and 0 otherwise.

Ideas 2: If the number has qualified a number, and it appears even more than the number of all the other numbers appear.
Save when traversing an array of two values: one is a numeric array, one number. When traversing the next number, if it is previously stored in the same figure, plus the number 1 or decremented by 1; if the number is 0, the next number is saved, and the number is set to 1. After traversing the saved numbers is also desired. And then determine whether it meets the conditions can be.
This actually is a "fractal leaf" method, the array simultaneously remove two different numbers, the last remaining of a number is the number. If the remaining two, then the two is the same, is the result, on the basis of its last remaining digit or two back to the original array, the array traversal again statistics about the numbers of occurrences for final judgment.

2, programming

python

Code implementation:

1 idea:

# -*- coding:utf-8 -*-
class Solution:
    def MoreThanHalfNum_Solution(self, numbers):
        # write code here
        numbers.sort()
        theone = numbers[len(numbers)/2]
        if numbers.count(theone) > len(numbers)/2:
            return theone
        return  0

2 ideas:

# -*- coding:utf-8 -*-
class Solution:
    def MoreThanHalfNum_Solution(self, numbers):
        # write code here
        if not numbers:
            return 0
        res = numbers[0]
        times = 1
        length = len(numbers)
        for i in range(1, length):
            if times == 0:
                res = numbers[i]
                times = 1
            elif res == numbers[i]:
                times += 1
            else:
                times -= 1
    
        import collections
        return res if collections.Counter(numbers)[res] * 2 > length else 0

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