今天再练一道迭代搜索的题 + 力扣每日一题

Power Calculus

题目描述

给你一个整数n，可以乘或者除，可以使用中间结果，求最少运算多少次可以得到$x^n$

分析

相当于求从整数1开始，可以加或者减，可以使用中间结果，最少运算多少次可以得到n

实现细节见代码

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;const int N = 1010; 
int n;
int ans[N]; // 保存中间结果，n最大为1000，每次运算加1，也就需要1000次，所以最多运算1000次，也就是说中间结果最多有1000个
int pos; // 标记当前运算的位置
int depth; // 最大深度// dep：当前递归深度
bool dfs(int dep) {if (dep > depth) return false;if (ans[pos] << (depth - dep) < n) {// depth - dep是还剩的递归次数// 估价函数：用最快的倍增都不能到达n，退出// 最快的倍增：假设ans[pos] = m, 下一层递归2m，再下一层4m， 8m 。。。 ,以2的幂倍增return false;}if (ans[pos] == n) {return true;}pos++;for (int i = 0; i < pos; ++i) {// 讨论上一个中间结果与前面所有的中间结果相加ans[pos] = ans[pos - 1] + ans[i];if (dfs(dep + 1)) {// 继续下一层的递归return true;}// 讨论上一个中间结果与前面所有的中间结果相减ans[pos] = ans[pos - 1] - ans[i];if (dfs(dep + 1)) {return true;}}pos--; // 回溯return false;
}int main() {while (scanf("%d", &n)) {if (n == 0) break;depth = 0;pos = 0;ans[pos] = 1;while (!dfs(0)) {depth++;}printf("%d\n", depth);}return 0;
}

706. 设计哈希映射

方法一：纯数组

class MyHashMap {private static final int MAX_N = 1000001;int[] map;public MyHashMap() {map = new int[MAX_N];Arrays.fill(map, -1);}public void put(int key, int value) {map[key] = value;}public int get(int key) {return map[key];}public void remove(int key) {map[key] = -1;}
}/*** Your MyHashMap object will be instantiated and called as such:* MyHashMap obj = new MyHashMap();* obj.put(key,value);* int param_2 = obj.get(key);* obj.remove(key);*/

方法二：数组加链表

class MyHashMap {static class Node {int key, value;Node next;Node(int key, int value) {this.key = key;this.value = value;}}public Node[] table = new Node[(int)Math.pow(2, 14)];public MyHashMap() {}public int getIndex(int key) {int hash = Integer.hashCode(key);hash = hash ^ (hash >>> 16);return hash & (table.length - 1);}public void put(int key, int value) {int index = getIndex(key);if (table[index] == null) {table[index] = new Node(key, value);} else {Node cur = table[index];Node pre = null;while (cur != null) {if (cur.key == key) {cur.value = value;return;}pre = cur;cur = cur.next;}Node node = new Node(key, value);pre.next = node;}}public int get(int key) {int index = getIndex(key);if (table[index] != null) {Node cur = table[index];while (cur != null) {if (cur.key == key) {return cur.value;}cur = cur.next;}}return -1;}public void remove(int key) {int index = getIndex(key);if (table[index] != null) {Node pre = null;Node cur = table[index];while (cur != null) {if (cur.key == key) {if (pre == null) {table[index] = cur.next;} else {pre.next = cur.next;}return;}pre = cur;cur = cur.next;}}}
}