Create (Contest 352) 2761. Prime Pairs With Target Sum.java

Author: ChenHCY

Array - Prime number/(Contest 352) 2761. Prime Pairs With Target Sum.java

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+/*Leetcode (Contest 352)  2761. Prime Pairs With Target Sum
+
+You are given an integer n. We say that two integers x and y form a prime number pair if:
+
+      · 1 <= x <= y <= n
+      · x + y == n
+      · x and y are prime numbers
+
+Return the 2D sorted list of prime number pairs [xi, yi]. The list should be sorted in increasing order of xi. If there are no prime number pairs at all, return an empty array.
+
+Note: A prime number is a natural number greater than 1 with only two factors, itself and 1.
+
+Example 1:
+
+Input: n = 10
+Output: [[3,7],[5,5]]
+Explanation: In this example, there are two prime pairs that satisfy the criteria. 
+These pairs are [3,7] and [5,5], and we return them in the sorted order as described in the problem statement.
+
+Example 2:
+
+Input: n = 2
+Output: []
+Explanation: We can show that there is no prime number pair that gives a sum of 2, so we return an empty array. 
+ 
+
+Constraints:
+1 <= n <= 10^6
+*/
+
+class Solution {
+    /*埃拉托斯特尼筛法（埃氏筛）是一种用于枚举质数的算法。
+     它的基本思想是从2开始，将每个素数的倍数标记为非素数，直到遍历完所有小于给定上限的数。
+
+     质数：质数是大于 1 的自然数，并且只有两个因子，即它 本身 和 1 。
+    */
+
+    private final static int MAX = (int) 1e6;
+    private final static HashSet<Integer> set = new HashSet<>(); //用来储存范围内的所有质数
+    static{
+        boolean[] isPrime = new boolean[MAX + 1];
+        //2是质数，则将其后面2的倍数的数字删除，因为他们都是2的倍数，所以不可能是质数，对于3，则删除3的倍数，
+        //==> 以此类推，后面无法删除的数字则是质数
+        for(int i = 2; i <= MAX; i++){
+            if(!isPrime[i]){
+                set.add(i);// 把得到的质数 加入到Hashset中储存
+                for(int j = i; j <= MAX / i; j++){
+                    isPrime[j * i] = true; //如果存在新的因数，则标记true
+                }
+            }
+        }
+    }
+    
+    //对于一个数字i，若i为质数且n-i也为质数，则这两个数字为满足题目要求的一个质数对
+    //那么只需要在n/2的一半范围内求解即可
+    public List<List<Integer>> findPrimePairs(int n) {
+        List<List<Integer>> res = new ArrayList<>();
+        for(int i = 2; i <= n/2; i++){
+            if(set.contains(i) && set.contains(n - i) && i <= n - i){
+                res.add(Arrays.asList(i, n - i));
+            }
+        }
+        return res;
+    }
+}