贝壳找房4月12日算法笔试题AC代码(1,2,3)
注 : 本贴子中的图片来自 https://www.nowcoder.com/discuss/638546?type=post&order=create&pos=&page=1&channel=-1&source_id=search_post_nctrack
1. 给定整数数组 A(A 中元素取值 0~9), 每次从 A 的首部或尾部选择一个数加入新的队列直到 A 为空, 问新的队列组成的数(从左往右看)最小是多少(可以包含前导 0)
#include <iostream> #include <vector> using namespace std; bool cmp(vector<int>& a, int l, int r){ // 通过比较函数来确定加入新队列的元素 while (l < r){ if (a[l] < a[r]) return true; else if (a[l] == a[r]) { ++l; --r; } else return false; } return true; } int main(){ int n; cin >> n; vector<int> a(n); for (int i = 0; i < n; i++){ cin >> a[i]; } vector<int> v; int i = 0, j = n - 1; while (i <= j){ if (cmp(a, i, j)) v.push_back(a[i++]); else v.push_back(a[j--]); } for (int k = 0; k < n; k++){ cout << v[k]; } cout << endl; return 0; }2.
用双向链表进行模拟
#include <iostream> #include <vector> using namespace std; const int INF = 1000000000; struct ListNode{ int val; ListNode *prev, *next; ListNode(int x) : val(x), prev(NULL), next(NULL) {} ListNode(int x, ListNode* p, ListNode* q) : val(x), pre***ext(q) {} }; void constructList(ListNode* head, int n){ ListNode* p = head; int x; for (int i = 0; i < n; i++){ scanf("%d", &x); p->next = new ListNode(x, p, NULL); p = p->next; } } void run(ListNode* head){ ListNode *p; int mx = INF; for (p = head->next; p != NULL; p = p->next){ if (p->val < mx){ printf("%d ", p->val); mx = p->val; p->prev->next = p->next; if (p->next != NULL){ p->next->prev = p->prev; } } } printf("\n"); } int main(){ int n; scanf("%d", &n); ListNode* head = new ListNode(0); constructList(head, n); while (head->next != NULL){ run(head); } return 0; }3.
标准的回溯算法。由于要输出最终字典序最小的方案, 用一个数组来记录。
#include <iostream> #include <vector> #include <algorithm> using namespace std; int ans = 100; vector<int> order; int n, m; vector<int> a; vector<vector<int>> b; bool check(vector<int>& sum){ for (int i = 0; i < n; i++){ if (sum[i] < a[i]) return false; } return true; } void dfs(vector<int>& sum, vector<int>& arr, int cnt, int cur){ if (check(sum) && cnt < ans){ ans = cnt; order.clear(); order.assign(arr.begin(), arr.end()); return; } if (cur == m) return; // chose b[cur] for (int i = 0; i < n; i++){ sum[i] += b[cur][i]; } arr.push_back(cur + 1); dfs(sum, arr, cnt + 1, cur + 1); arr.pop_back(); for (int i = 0; i < n; i++){ sum[i] -= b[cur][i]; } //abandon b[cur] dfs(sum, arr, cnt, cur + 1); } int main(){ cin >> n; a.resize(n); for (int i = 0; i < n; i++){ cin >> a[i]; } cin >> m; b.resize(m, vector<int>(n)); for (int i = 0; i < m; i++){ for (int j = 0; j < n; j++){ cin >> b[i][j]; } } vector<int> sum(n), arr; dfs(sum, arr, 0, 0); cout << ans << endl; for (int x : order) cout << x << ' '; cout << endl; return 0; }4.
不会做,也没想到方法。
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