Review: 4

Author: rchen102

Java/leetcode 187.java

@@ -14,37 +14,8 @@ public List<String> findRepeatedDnaSequences(String s) {
     }
 }
 
-//Solution2: bit manipulation and HashSet  T: O(n) S: O(n)
-class Solution {
-    public List<String> findRepeatedDnaSequences(String s) {
-        List<String> list = new ArrayList<>();
-        if(s.length() < 10) return list;
-        
-        Set<Integer> set = new HashSet<>();
-        Set<Integer> repeated = new HashSet<>();
-        
-        //coding of a 10-letter sequence into a 4-bytes number
-        char[] map = new char[26];
-        //A:00 
-        map['C'-'A'] = 1; //C:01 
-        map['G'-'A'] = 2; //G:10 
-        map['T'-'A'] = 3; //T:11 
-        
-        for(int i = 0; i < s.length() - 9; i++) {
-            int v = 0;
-            for(int j = i; j < i + 10; j++) {
-                v <<= 2;
-                v |= map[s.charAt(j) - 'A'];
-            }
-            if(!set.add(v) && repeated.add(v)) {
-                list.add(s.substring(i, i + 10));
-            }
-        }
-        return list;
-    }
-}
-
-//Solution2: follow up 
+// Solution2: bit manipulation and HashSet  
+// T: O(n) S: O(n)
 class Solution {
     public List<String> findRepeatedDnaSequences(String s) {
         List<String> list = new ArrayList<>();
@@ -64,7 +35,7 @@ public List<String> findRepeatedDnaSequences(String s) {
         for(int i = 0; i < s.length(); i++) {
             v <<= 2;
             v |= map[s.charAt(i) - 'A'];
-            v &= 0xfffff;
+            v &= (1 << 20) - 1;
             if(i < 9)
                 continue;
             if(!set.add(v) && repeated.add(v)) {

Java/leetcode 205.java

@@ -21,7 +21,8 @@ public boolean isIsomorphic(String s, String t) {
     }
 }
 
-//Own Solution: HashMap T: O(n) S: O(512)
+//Own Solution: HashMap 
+//T: O(n) S: O(512)
 class Solution {
     public boolean isIsomorphic(String s, String t) {
         Map<Character, Integer> mapS = new HashMap<>();

Java/leetcode 215.java

@@ -55,16 +55,34 @@ private void swap(int[] nums, int i, int j) {
     }
 }
 
-// Solution2: priorityqueue T: O(nlogk) S: O(k)
+// Solution2: Heap(PriorityQueue)
+// T: O(nlogk) S: O(k)
 class Solution {
     public int findKthLargest(int[] nums, int k) {
+        if (nums == null || nums.length == 0 ) return -1;
         PriorityQueue<Integer> pq = new PriorityQueue<>();
-        for(int num : nums) {
+        for (int num : nums) {
+            if (pq.size() >= k) {
+                if (pq.peek() < num) {
+                    pq.poll();
+                    pq.offer(num);
+                }
+            }
+            else pq.offer(num);
+        }
+        return pq.poll();
+    }
+
+    // 简化，复杂度稍高，会增加不必要的插入、删除操作
+    public int findKthLargest(int[] nums, int k) {
+        if (nums == null || nums.length == 0 ) return -1;
+        PriorityQueue<Integer> pq = new PriorityQueue<>();
+        for (int num : nums) {
             pq.offer(num);
-            if(pq.size() > k) {
+            if (pq.size() > k) {
                 pq.poll();
             }
         }
-        return pq.peek();
+        return pq.poll();
     }
 }

Java/leetcode 49.java

@@ -1,13 +1,11 @@
-// Solution1: HashMap + sort
-// T: O(n * klogk) S: O(n)
+// Solution1: HashMap + encoding function 
+// T: O(n * k) S: O(n)
 class Solution {
     public List<List<String>> groupAnagrams(String[] strs) {
         if (strs == null || strs.length == 0) return null;
         Map<String, List<String>> map = new HashMap<>();
         for (String str : strs) {
-            char[] chs = str.toCharArray();
-            Arrays.sort(chs);
-            String tag = String.valueOf(chs);
+            String tag = this.encode(str);
 
             if (!map.containsKey(tag)) {
                 map.put(tag, new ArrayList<>());
@@ -16,16 +14,26 @@ public List<List<String>> groupAnagrams(String[] strs) {
         }
         return new ArrayList<>(map.values());
     }
+    
+    private String encode(String str) {
+        char[] code = new char[26];      // ASCLL = 0 by default
+        for (char ch : str.toCharArray()) {
+            code[ch-'a'] += 1;
+        }
+        return String.valueOf(code);
+    }
 }
 
-// Solution2: HashMap + encoding function 
-// T: O(n * k) S: O(n)
+// Solution2: HashMap + sort
+// T: O(n * klogk) S: O(n)
 class Solution {
     public List<List<String>> groupAnagrams(String[] strs) {
         if (strs == null || strs.length == 0) return null;
         Map<String, List<String>> map = new HashMap<>();
         for (String str : strs) {
-            String tag = this.encode(str);
+            char[] chs = str.toCharArray();
+            Arrays.sort(chs);
+            String tag = String.valueOf(chs);
 
             if (!map.containsKey(tag)) {
                 map.put(tag, new ArrayList<>());
@@ -34,12 +42,4 @@ public List<List<String>> groupAnagrams(String[] strs) {
         }
         return new ArrayList<>(map.values());
     }
-    
-    private String encode(String str) {
-        char[] code = new char[26];      // ASCLL = 0 by default
-        for (char ch : str.toCharArray()) {
-            code[ch-'a'] += 1;
-        }
-        return String.valueOf(code);
-    }
 }