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RMQ_02.java
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104 lines (92 loc) · 3.09 KB
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package _aTemplate;
// 小红书
// 3.13 笔试
// 给定一个数组，想随时查询任何范围上的最大值
// 如果只是根据初始数组建立、并且以后没有修改，
// 那么RMQ方法比线段树方法好实现，时间复杂度O(N*logN)，额外空间复杂度O(N*logN)
public class RMQ_02 {
public static class RMQ {
public int[][] max;
public RMQ(int[] arr) {
int n = arr.length;
int k = power2(n);
max = new int[n + 1][k + 1]; // 下标从1开始
for (int i = 1; i <= n; i++) {
max[i][0] = arr[i - 1];
}
for (int j = 1; (1 << j) <= n; j++) {
for (int i = 1; i + (1 << j) - 1 <= n; i++) {
max[i][j] = Math.max(
max[i][j - 1],
max[i + (1 << (j - 1))][j - 1]);
}
}
}
public int max(int l, int r) {
// l...r -> r - l + 1 -> 2的哪个次方最接近它！
int k = power2(r - l + 1);
return Math.max(max[l][k], max[r - (1 << k) + 1][k]);
}
// 离n最近的2的某次幂
private int power2(int n) {
int ans = 0;
// m >> 1 : m先减半为了退出循环时 ans 不用--
while ((1 << ans) <= (n >> 1)) {
ans++;
}
return ans;
}
}
// 为了测试
public static class Right {
public int[][] max;
public Right(int[] arr) {
int n = arr.length;
max = new int[n + 1][n + 1];
for (int l = 1; l <= n; l++) {
max[l][l] = arr[l - 1];
for (int r = l + 1; r <= n; r++) {
max[l][r] = Math.max(max[l][r - 1], arr[r - 1]);
}
}
}
public int max(int l, int r) {
return max[l][r];
}
}
// 为了测试
public static int[] randomArray(int n, int v) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = (int) (Math.random() * v);
}
return arr;
}
// 为了测试
public static void main(String[] args) {
int N = 150;
int V = 200;
int testTimeOut = 20000;
int testTimeIn = 200;
System.out.println("测试开始");
for (int i = 0; i < testTimeOut; i++) {
int n = (int) (Math.random() * N) + 1;
int[] arr = randomArray(n, V);
int m = arr.length;
RMQ rmq = new RMQ(arr);
Right right = new Right(arr);
for (int j = 0; j < testTimeIn; j++) {
int a = (int) (Math.random() * m) + 1;
int b = (int) (Math.random() * m) + 1;
int l = Math.min(a, b);
int r = Math.max(a, b);
int ans1 = rmq.max(l, r);
int ans2 = right.max(l, r);
if (ans1 != ans2) {
System.out.println("出错了!");
}
}
}
System.out.println("测试结束");
}
}