Leetcode Maximum XOR of Two Numbers in an Array

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题意：给出一列数组，求出其中两个数字异或的最大值。

思路：这题的思路很巧妙，枚举了每个位上的数字。又利用的异或的性质。

class Solution {public:    int findMaximumXOR(vector<int>& nums) {        int mask = 0;        int re = 0;        for(int i = 31; i >=0; -- i) {            mask |= (1 << i);            map<int, bool> prefix;            for(int j = 0; j < nums.size(); ++ j) {                prefix[mask & nums[j]] = true;            }                        int temp = re;            temp |= (1 << i);            for(int j = 0; j < nums.size(); ++ j) {                if(prefix[temp ^ (nums[j] & mask)]) re = temp;            }        }                return re;    }};

当查询某个数是否存在时，可以用字典树，而不是用map。

class Solution {public:    int findMaximumXOR(vector<int>& nums) {        TrieTree* root = buildTrieTree(nums);        //root = root->buildTrieTree(nums);        int re = 0;        for(int i = 0; i < nums.size(); ++ i) {            re = max(re, findIt(root, nums[i]));        }                return re;    }    private:    class TrieTree {    public:        TrieTree* c[2];        TrieTree() {            c[0] = NULL, c[1] = NULL;        };    };        TrieTree* buildTrieTree(vector<int> nums) {            TrieTree* root = new TrieTree(), *next = NULL;            for(int i = 0; i < nums.size(); ++ i) {                int num = nums[i]; //cout << num << endl;                next = root;                                for(int j = 31; j >=0; --j) {                    bool bit = (num >> j) & 1; //cout <<num << ((num >>= j) & 1) << endl;                    if(next->c[bit]) {                        next = next->c[bit];                    }                    else {                        TrieTree* child = new TrieTree();                        next->c[bit] = child;                        next = child;                    }                }            }                        return root;        }        int findIt(TrieTree* root, int num) {        int re = 0;        int rnum = ~num;                for(int i = 31; i >=0; -- i) {            bool bit =(rnum >> i) & 1;            if(root->c[bit]) {                re |= (1 << i);                root = root->c[bit];            }            else {                root = root->c[!bit];            }        }                return re;    }};

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