力扣 1707. 与数组中元素的最大异或值 01字典树 贪心

https://leetcode-cn.com/problems/maximum-xor-with-an-element-from-array/

思路：01字典树，只不过多了一个最大值限制。那么在字典树的每个节点中记录一下到当前位置的最小值，如果该最小值满足限制，那么就可以走到该节点，否则不行。

class Tire{
public:
    vector<Tire*> child;
    int min_value;

    Tire():min_value(0x3f3f3f3f)
    {
        child.resize(2,nullptr);
    }

    void insert(int val)
    {
        int bin_value=1<<30;
        min_value=min(min_value,val);
        Tire *cur=this;
        while(bin_value)
        {
            int idx=min(1,val&bin_value);
            if(cur->child[idx]==nullptr)
                cur->child[idx]=new Tire();
            cur=cur->child[idx];
            cur->min_value=min(cur->min_value,val);
            bin_value>>=1;
        }
    }

    int query(int val,int limit)
    {
        int bin_value=1<<30;
        int ans=0;
        Tire *cur=this;
        if(cur->min_value>limit)
            return -1;
        while(cur)
        {
            int idx=min(1,val&bin_value);
            if(cur->child[1-idx]&&cur->child[1-idx]->min_value<=limit)
            {
                ans|=bin_value;
                cur=cur->child[1-idx];
            }
            else
                cur=cur->child[idx];
            bin_value>>=1;
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Tire root;
        for(const int& num:nums)
            root.insert(num);
        vector<int> ans;
        for(const vector<int>& pr:queries)
            ans.push_back(root.query(pr[0],pr[1]));
        return ans;
    }
};