LeetCode #Single Number#

解法一:

    人类需要O(n)去解决问题，于是普罗米修斯不管三七二十一就偷来了Hash...    Python里面内置的dic好用到不行．这里可以利用Hash把时间复杂度降到O(n)，但是这种解法不满足对内存的要求．．．

"""Programmer  :   EOFCode date   :   2015.03.02file        :   sn.pye-mail      :   [e-mail] description :　　　　　　　Given an array of integers, every element appears twice except for one. Find that single one.Note:    Your algorithm should have a linear runtime complexity.Could you implement it without using extra memory?In this solution, I used a hash/dictionary table."""class Solution():    def singleNumber(self, A) :        dic = {}        for i in range(0, len(A)) :            dic[A[i]] = 0        for i in range(0, len(A)) :            if A[i] in dic :                dic[A[i]] += 1        for i in range(0, len(A)) :            if dic[A[i]] == 1 :                return A[i]#----------for testing ----------------array = [1,2,3,4,5,4,3,2,1]s = Solution()print s.singleNumber(array)

解法二：

首先最简单的思路不过就是排序，一旦排序起来，只要有数值的，就最好不要用比较排序，基于比较排序最快也就O(n * log(n) )，线性时间排序O(n)不能再sexy...　但是这种方法不满足对内存的要求(Don't use extra memory).

"""Programmer  :   EOFCode date   :   2015.03.02file        :   sn.pye-mail      :   [e-mail] description :　　　　　　　Given an array of integers, every element appears twice except for one. Find that single one.Note:    Your algorithm should have a linear runtime complexity.Could you implement it without using extra memory?    In this solution, I sort the inputed @array by @counting sort which is a very fast algorithm which's runtime complexityis O(n). BUT I have to use extra memory to solve this problem bythis method."""class Solution():    def counting_sort(self, array) :        array_size = len(array)        max_value = 0        for i in range(0, array_size) :            if max_value < array[i] :                max_value = array[i]        buf_size = max_value+1        buf = [0] * buf_size        ret = [0] * array_size        for i in range(0, array_size) :            buf[array[i]] += 1        for i in range(1, buf_size) :            buf[i] += buf[i-1]        for i in range(array_size-1, -1, -1) :            ret[buf[array[i]] - 1] = array[i]            buf[array[i]] -= 1        return ret    def singleNumber(self, A) :        A = self.counting_sort(A)        length = len(A)        for i in range(0, length, 2) :            if i < length-2 and A[i] != A[i+1] :                return A[i]            elif i == length-1 :                return A[i]#----------for testing ----------------array = [1,2,3,4,5,4,3,2,1]s = Solution()print s.singleNumber(array)

解法三:

这种解法是我看到 @Powerxing　的blog时候看到的．利用．．．异或

节操在这里:http://www.powerxing.com/leetcode-single-number/

"""Programmer  :   EOFCode date   :   2015.03.02file        :   sn3.pye-mail      :   [e-mail] description :    Given an array of integers, every element appears twice except for one. Find that single one.Note:    Your algorithm should have a linear runtime complexity.Could you implement it without using extra memory?In this solution, I used a XOR operand to solve this problem."""class Solution():    def singleNumber(self, A) :        result = 0        for i in range(0, len(A)) :            result ^= A[i]        return result#----------for testing ----------------array = [1,2,3,4,5,4,3,2,1]s = Solution()print s.singleNumber(array)