update

Author: OneCodeMonkey

leetcode_solved/[editing]leetcode_0217_Contains_Duplicate.cpp

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+// AC：T:O(n), S:O(n)
+class Solution {
+    public boolean containsDuplicate(int[] nums) {
+        HashSet<Integer> record = new HashSet<>();
+        int size = nums.length;
+        for(int i = 0; i < size; i++) {
+            if (record.contains(nums[i])) {
+                return true;
+            }
+            record.add(nums[i]);
+        }
+        return false;
+    }
+}
+
+// 法二：先排序，再判断连续两个元素是否相等。
+// T:O(n logn), S:O(logn) (调用栈深度)

leetcode_solved/[editing]leetcode_0219_Contains_Duplicate_II.cpp

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+/**
+ * AC: Runtime: 5 ms, faster than 91.91% of Java online submissions for Contains Duplicate II.
+ * Memory Usage: 45 MB, less than 30.98% of Java online submissions for Contains Duplicate II.
+ * T: O(n), S: O(n)
+ * hashmap 存某个元素最近一次出现的位置，和当前如果相同的元素比较索引，如果之差绝对值 <= k 即满足返回。
+ */
+class Solution {
+    public boolean containsNearbyDuplicate(int[] nums, int k) {
+        HashMap<Integer, Integer> record = new HashMap<Integer, Integer>();
+        int size = nums.length;
+        for(int i = 0; i < size; i++) {
+            if (record.get(nums[i]) != null) {
+                if (i - record.get(nums[i]) <= k) {
+                    return true;
+                }
+            }
+            record.put(nums[i], i);
+        }
+        return false;
+    }
+}