Contains Duplicate I & II

Author: int32bit

README.md

@@ -122,6 +122,8 @@
 + [211 Add and Search Word - Data structure design(Trie树)](algorithms/AddandSearchWord)
 + [213 House Robber II(DP)](algorithms/HouseRobberII)
 + [215 Kth Largest Element in an Array(快排partition函数)](algorithms/KthLargestElementinanArray)
++ [217 Contains Duplicate(hashset)](algorithms/ContainsDuplicate)
++ [219 Contains Duplicate II(hashmap)](algorithms/ContainsDuplicateII)
 
 ## Database
 

algorithms/ContainsDuplicate/README.md

@@ -0,0 +1,26 @@
+## Contains Duplicate
+
+Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.
+
+## Solution
+
+直接使用hashtable即可
+
+```cpp
+bool containsDuplicate(vector<int> &nums) {
+	unordered_set<int> set;
+	for (int i : nums) {
+		if (set.find(i) != set.end())
+			return true;
+		else
+			set.insert(i);
+	}
+	return false;
+}
+```
+
+## 扩展
+
+[Contains Duplicate](../ContainsDuplicate): 判断一个数组是否存在重复元素
+[Contains Duplicate II](../ContainsDuplicateII):判断一个数组在指定距离内是否存在重复元素
+[Contains Duplicate III](../ContainsDuplicateIII):判断一个数组在指定距离内是否存在元素距离小于t

algorithms/ContainsDuplicate/solve.cpp

@@ -0,0 +1,25 @@
+#include <vector>
+#include <unordered_set>
+#include <string>
+#include <iostream>
+using namespace std;
+class Solution {
+	public:
+		bool containsDuplicate(vector<int> &nums) {
+			unordered_set<int> set;
+			for (int i : nums) {
+				if (set.find(i) != set.end())
+					return true;
+				else
+					set.insert(i);
+			}
+			return false;
+		}
+};
+int main(int argc, char **argv)
+{
+	Solution solution;
+	vector<int> nums = {2,3,4,1};
+	cout << solution.containsDuplicate(nums) << endl;
+	return 0;
+}

algorithms/ContainsDuplicateII/README.md

@@ -0,0 +1,29 @@
+## Contains Duplicate II
+
+Given an array of integers and an integer k, find out whether there there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k.
+
+## Solution
+
+使用hashmap，key存储值，value存储索引, 若`nums[i]`在hashmap中，则若`i - map[nums[i]] <= k`, 返回true
+
+```cpp
+bool containsNearbyDuplicate(vector<int> &nums, int k) {
+	unordered_map<int, int> map;
+	int n = nums.size();
+	for (int i = 0; i < n; ++i) {
+		if (map.find(nums[i]) != map.end()) {
+			int s = map[nums[i]];
+			if (i - s <= k)
+				return true;
+		}
+		map[nums[i]] = i;
+	}
+	return false;
+}
+```
+
+## 扩展
+
+[Contains Duplicate](../ContainsDuplicate): 判断一个数组是否存在重复元素
+[Contains Duplicate II](../ContainsDuplicateII):判断一个数组在指定距离内是否存在重复元素
+[Contains Duplicate III](../ContainsDuplicateIII):判断一个数组在指定距离内是否存在元素距离小于t

algorithms/ContainsDuplicateII/solve.cpp

@@ -0,0 +1,28 @@
+#include <vector>
+#include <unordered_map>
+#include <string>
+#include <iostream>
+using namespace std;
+class Solution {
+	public:
+		bool containsNearbyDuplicate(vector<int> &nums, int k) {
+			unordered_map<int, int> map;
+			int n = nums.size();
+			for (int i = 0; i < n; ++i) {
+				if (map.find(nums[i]) != map.end()) {
+					int s = map[nums[i]];
+					if (i - s <= k)
+						return true;
+				}
+				map[nums[i]] = i;
+			}
+			return false;
+		}
+};
+int main(int argc, char **argv)
+{
+	Solution solution;
+	vector<int> nums = {2,3,4,1};
+	cout << solution.containsNearbyDuplicate(nums, 1) << endl;
+	return 0;
+}