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package class06;
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// 测试链接 : https://leetcode.com/problems/maximum-xor-with-an-element-from-array/
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public class Code03_MaximumXorWithAnElementFromArray {
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public static int[] maximizeXor(int[] nums, int[][] queries) {
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int N = nums.length;
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NumTrie trie = new NumTrie();
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for (int i = 0; i < N; i++) {
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trie.add(nums[i]);
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}
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int M = queries.length;
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int[] ans = new int[M];
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for (int i = 0; i < M; i++) {
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ans[i] = trie.maxXorWithXBehindM(queries[i][0], queries[i][1]);
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}
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return ans;
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}
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public static class Node {
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public int min;
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public Node[] nexts;
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public Node() {
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min = Integer.MAX_VALUE;
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nexts = new Node[2];
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}
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}
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public static class NumTrie {
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public Node head = new Node();
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public void add(int num) {
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Node cur = head;
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head.min = Math.min(head.min, num);
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for (int move = 30; move >= 0; move--) {
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int path = ((num >> move) & 1);
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cur.nexts[path] = cur.nexts[path] == null ? new Node() : cur.nexts[path];
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cur = cur.nexts[path];
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cur.min = Math.min(cur.min, num);
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}
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}
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// 这个结构中,已经收集了一票数字
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// 请返回哪个数字与X异或的结果最大,返回最大结果
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// 但是,只有<=m的数字,可以被考虑
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public int maxXorWithXBehindM(int x, int m) {
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if (head.min > m) {
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return -1;
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}
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// 一定存在某个数可以和x结合
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Node cur = head;
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int ans = 0;
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for (int move = 30; move >= 0; move--) {
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int path = (x >> move) & 1;
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// 期待遇到的东西
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int best = (path ^ 1);
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best ^= (cur.nexts[best] == null || cur.nexts[best].min > m) ? 1 : 0;
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// best变成了实际遇到的
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ans |= (path ^ best) << move;
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cur = cur.nexts[best];
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}
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return ans;
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}
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}
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}
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