From 06ebd5f204f8a7012c1e4518ce778fa06750d7e6 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=93=88=E5=93=88=E5=93=88?=
 <76643786+Projecthappy@users.noreply.github.com>
Date: Thu, 23 Mar 2023 00:08:57 +0800
Subject: [PATCH] =?UTF-8?q?Update=200454.=E5=9B=9B=E6=95=B0=E7=9B=B8?=
 =?UTF-8?q?=E5=8A=A0II.md?=
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原先是先判断map中是否有这个key，后面更新时再get(key)，经过测试，直接使用getOrDefault()方法效率会更高一些，没有当前key时返回0，值为0就是没有当前key
---
 problems/0454.四数相加II.md | 17 ++++++++---------
 1 file changed, 8 insertions(+), 9 deletions(-)

diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index c2a5710f..411b60e8 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -97,26 +97,25 @@ Java：
 ```Java
 class Solution {
     public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
-        Map<Integer, Integer> map = new HashMap<>();
-        int temp;
         int res = 0;
+        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
         //统计两个数组中的元素之和，同时统计出现的次数，放入map
         for (int i : nums1) {
             for (int j : nums2) {
-                temp = i + j;
-                if (map.containsKey(temp)) {
-                    map.put(temp, map.get(temp) + 1);
+                int tmp = map.getOrDefault(i + j, 0);
+                if (tmp == 0) {
+                    map.put(i + j, 1);
                 } else {
-                    map.put(temp, 1);
+                    map.replace(i + j, tmp + 1);
                 }
             }
         }
         //统计剩余的两个元素的和，在map中找是否存在相加为0的情况，同时记录次数
         for (int i : nums3) {
             for (int j : nums4) {
-                temp = i + j;
-                if (map.containsKey(0 - temp)) {
-                    res += map.get(0 - temp);
+                int tmp = map.getOrDefault(0 - i - j, 0);
+                if (tmp != 0) {
+                    res += tmp;
                 }
             }
         }