From d01fd238549bc10a6bd5b883f91d60d1ed2ab419 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Fri, 3 Sep 2021 13:23:03 +0800
Subject: [PATCH] =?UTF-8?q?454:=E5=9B=9B=E6=95=B0=E7=9B=B8=E5=8A=A0=20II?=
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---
 .../java/leetcode/editor/cn/FourSumIi.java    | 62 +++++++++++++++++++
 .../editor/cn/doc/content/FourSumIi.md        | 22 +++++++
 2 files changed, 84 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/FourSumIi.java
 create mode 100644 src/main/java/leetcode/editor/cn/doc/content/FourSumIi.md

diff --git a/src/main/java/leetcode/editor/cn/FourSumIi.java b/src/main/java/leetcode/editor/cn/FourSumIi.java
new file mode 100644
index 0000000..0631444
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/FourSumIi.java
@@ -0,0 +1,62 @@
+//给定四个包含整数的数组列表 A , B , C , D ,计算有多少个元组 (i, j, k, l) ，使得 A[i] + B[j] + C[k] + D[
+//l] = 0。 
+//
+// 为了使问题简单化，所有的 A, B, C, D 具有相同的长度 N，且 0 ≤ N ≤ 500 。所有整数的范围在 -2²⁸ 到 2²⁸ - 1 之间，最
+//终结果不会超过 2³¹ - 1 。 
+//
+// 例如: 
+//
+// 
+//输入:
+//A = [ 1, 2]
+//B = [-2,-1]
+//C = [-1, 2]
+//D = [ 0, 2]
+//
+//输出:
+//2
+//
+//解释:
+//两个元组如下:
+//1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
+//2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
+// 
+// Related Topics 数组 哈希表 👍 408 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.HashMap;
+import java.util.Map;
+
+//454:四数相加 II
+class FourSumIi {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new FourSumIi().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
+            Map<Integer, Integer> map1 = new HashMap<>();
+            Map<Integer, Integer> map2 = new HashMap<>();
+            int size = nums1.length;
+            for (int i = 0; i < size; i++) {
+                for (int j = 0; j < size; j++) {
+                    map1.put(nums1[i] + nums2[j], map1.getOrDefault(nums1[i] + nums2[j], 0) + 1);
+                    map2.put(nums3[i] + nums4[j], map2.getOrDefault(nums3[i] + nums4[j], 0) + 1);
+                }
+            }
+            int count = 0;
+            for (int key : map1.keySet()) {
+                if (map2.containsKey(-key)) {
+                    count += map1.get(key) * map2.get(-key);
+                }
+            }
+            return count;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/doc/content/FourSumIi.md b/src/main/java/leetcode/editor/cn/doc/content/FourSumIi.md
new file mode 100644
index 0000000..6c7f496
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/FourSumIi.md
@@ -0,0 +1,22 @@
+<p>给定四个包含整数的数组列表&nbsp;A , B , C , D ,计算有多少个元组 <code>(i, j, k, l)</code>&nbsp;，使得&nbsp;<code>A[i] + B[j] + C[k] + D[l] = 0</code>。</p>
+
+<p>为了使问题简单化，所有的 A, B, C, D 具有相同的长度&nbsp;N，且 0 &le; N &le; 500 。所有整数的范围在 -2<sup>28</sup> 到 2<sup>28</sup> - 1 之间，最终结果不会超过&nbsp;2<sup>31</sup> - 1 。</p>
+
+<p><strong>例如:</strong></p>
+
+<pre>
+<strong>输入:</strong>
+A = [ 1, 2]
+B = [-2,-1]
+C = [-1, 2]
+D = [ 0, 2]
+
+<strong>输出:</strong>
+2
+
+<strong>解释:</strong>
+两个元组如下:
+1. (0, 0, 0, 1) -&gt; A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
+2. (1, 1, 0, 0) -&gt; A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
+</pre>
+<div><div>Related Topics</div><div><li>数组</li><li>哈希表</li></div></div><br><div><li>👍 408</li><li>👎 0</li></div>
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