添加阿里秋招2021

阿里2021秋招/排列组合.cpp

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+/*** 
+ * @Author: czqmike
+ * @Date: 2021-07-24 15:54:06
+ * @LastEditTime: 2021-07-24 16:48:25
+ * @LastEditors: czqmike
+ * @Description: 
+	给定一个n，求 [1,n] 这 n 个数字的排列组合有多少个。
+	条件：相邻的两个数字差的绝对值不能等于1.
+	例如：
+	4
+	[2, 4, 1, 3]
+	[3, 1, 4, 2]
+ * @FilePath: \Leetcode\阿里2021秋招\排列组合.cpp
+ */
+
+#include <iostream>
+#include <vector>
+#include <cmath>
+using namespace std;
+
+int n = 4;
+vector<vector<int>> res;
+vector<int> used(n + 1, 0);
+vector<int> temp(n, 0);
+
+void f(int index) { // index in [0, n - 1], stands layer.
+    if (index == n) {
+        res.push_back(temp);
+        return;
+    }
+
+    for (int i = 1; i <= n; ++i) { // i in [1, n], stands each number in layer.
+        if (!used[i] && (temp.size() == 0 || abs(i - temp[index - 1]) != 1)) {
+            // temp is empty or is not standing by.
+            used[i] = 1;
+            temp[index] = i;
+            f(index + 1);
+            used[i] = 0;
+        }
+    }
+}
+
+int main() {
+    f(0);
+
+    // cout << "res.size = " << res.size() << '\n';
+    for (auto v : res) {
+        for (auto item : v) {
+            cout << item << ' ';
+        }
+        cout << '\n';
+    }
+
+    return 0;
+}

阿里2021秋招/连续区间的数量.cpp

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+/*** 
+ * @Author: czqmike
+ * @Date: 2021-07-24 16:58:23
+ * @LastEditTime: 2021-07-24 18:24:19
+ * @LastEditors: czqmike
+ * @Description: 
+	长度为 n 的数组，数组中每个元素 a 满足：1<=a<=n
+	求连续区间的数量，要求区间中相同元素的数量 >=m
+	例如：
+	5 2
+	1 2 1 2 5
+	5
+	可以有5种可能：[1,3]，[1,4]，[1,5]，[2,5]，[1,4]
+	source: https://www.nowcoder.com/discuss/457491?channel=-2&source_id=discuss_terminal_discuss_sim
+ * @FilePath: \Leetcode\阿里2021秋招\连续区间的数量.cpp
+ */
+
+// 思路：用map存一个当前数的list集合，从前往后遍历，符合m条件的，则将对应数量添加到res
+// 并且维护一个变量pt，用来判断当前数不符合m条件，但之前有符合m条件的数出现过的首个数的下标
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int n = 0, m = 0;
+	int num;
+	int pt = -1, ans = 0;
+	unordered_map<int, vector<int>> hash; // the map means (number, number indexs vectors)
+
+	cin >> n >> m;
+	for (int i = 0; i < n; ++ i) {
+		cin >> num;
+		hash[num].push_back(i); // map::operator[] could access the map entry, but if the key does not exist, a new entry with that key will be created:
+		// Note that std::map::insert only succeeds when no existed keys in the map.
+
+		if (hash[num].size() >= m) {
+			int pre_m = hash[num].at(hash[num].size() - m); // get the index of biggest history pt.
+			pt = max(pt, pre_m);  // update pt.
+		}
+		ans += pt + 1; // not += pt, pt itself should be counted.
+	}
+
+	cout << ans << endl;
+    return 0;
+}
+