题目如下：

解题思路：本题的难点在于O(n)的复杂度。为了减少比较的次数，我们可以采用字典树保存输入数组中所有元素的二进制的字符串。接下来就是找出每个元素的异或的最大值，把需要找最大值的元素转成二进制表达后逐位在字典树中查找，查找的时候优先匹配反码，反码不存在则用原码。

代码如下：

class Solution(object):    def findMaximumXOR(self, nums):        """        :type nums: List[int]        :rtype: int        """        MAX_LEN = 31        root = {}        for i in nums:            node = root            i = '0'*(MAX_LEN-len(bin(i)[2:])) + bin(i)[2:]            for b in i:                if b not in node:                    node[b] = {}                node = node[b]        #print root['0']['0']['0']['1']['1']        res = 0        for i in nums:            path = ''            i = '0' * (MAX_LEN - len(bin(i)[2:])) + bin(i)[2:]            node = root            for b in i:                if len(node) == 0:                    break                ib = '1' if b == '0' else '0'                if ib in node:                    node = node[ib]                    path += ib                else:                    node = node[b]                    path += b            #print i,path,int(i,2),int(path,2)            res = max(res,int(i,2)^int(path,2))        return res