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Author: sc-yan

Java/Array/9-Maximum Product of Two Elements in an Array.java

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+Easy Leetcode 1464
+tags: Array
+
+方法1:
+题目不难。依旧是自己的思路，没有用任何数据结构。单纯array操作的话，效率最快
+方法2：
+priorityqueue。
+方法3：
+TreeMap
+
+
+```
+
+/*
+方法1
+*/
+//import java.util.stream.IntStream;
+    public int maxProduct(int[] nums) {
+        int max, max2nd; 
+        max = max2nd = Integer.MIN_VALUE;
+        for (int num : nums) {
+            if (num >= max) {
+                max2nd = max;
+                max = num;
+            } else if (num >= max2nd) {
+                max2nd = num;
+            }
+        }
+        return (max -1 ) * (max2nd -1);
+    }
+
+
+/*
+方法2
+*/
+ public int maxProduct(int[] nums) {
+        PriorityQueue<Integer> max = new PriorityQueue<>(Collections.reverseOrder());
+        for(int num: nums) max.offer(num);
+        // pop up the max number to get 2nd max
+        return (max.poll() -1) * (max.poll() -1);
+    }
+
+/*
+方法3
+*/
+public int maxProduct(int[] nums) {
+        TreeMap<Integer, Integer> map = new TreeMap<>();
+        // save how many times the number appears in the array. you will see why right now 
+        for (int i = 0; i < nums.length; i++) map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
+        if(map.get(map.lastKey()) >= 2) {
+            return (map.lastKey() -1) * (map.lastKey() -1);
+        }else{
+            // pop up the max number to get 2nd max
+            int max = map.lastKey() -1;
+            map.remove(map.lastKey());
+            return (max * (map.lastKey() -1));
+        }
+    }
+
+