From 1c871a9eb0d5a556b41409ac86750b878dd72e17 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E6=9C=B1=E5=8A=9B?= <171250568@smail.nju.edu.cn>
Date: Wed, 11 Nov 2020 14:41:25 +0800
Subject: [PATCH 1/3] =?UTF-8?q?=E3=80=90645.=20=E9=94=99=E8=AF=AF=E7=9A=84?=
=?UTF-8?q?=E9=9B=86=E5=90=88=E3=80=91=E3=80=90Java=E3=80=91?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
.../缺失和重复的元素.md | 30 +++++++++++++++++--
1 file changed, 28 insertions(+), 2 deletions(-)
diff --git a/高频面试系列/缺失和重复的元素.md b/高频面试系列/缺失和重复的元素.md
index 06e554f..52c8d78 100644
--- a/高频面试系列/缺失和重复的元素.md
+++ b/高频面试系列/缺失和重复的元素.md
@@ -1,4 +1,4 @@
-# 如何寻找缺失和重复的元素
+# 如何寻找缺失和重复的元素
@@ -139,4 +139,30 @@ vector findErrorNums(vector& nums) {
-======其他语言代码======
\ No newline at end of file
+======其他语言代码======
+
+[zhuli](https://github.com/1097452462 "zhuli")提供的Java代码：
+```java
+class Solution {
+ public int[] findErrorNums(int[] nums) {
+ int n = nums.length;
+ int dup = -1;
+ for (int i = 0; i < n; i++) {
+ // 元素是从 1 开始的
+ int index = Math.abs(nums[i]) - 1;
+ // nums[index] 小于 0 则说明重复访问
+ if (nums[index] < 0)
+ dup = Math.abs(nums[i]);
+ else
+ nums[index] *= -1;
+ }
+ int missing = -1;
+ for (int i = 0; i < n; i++)
+ // nums[i] 大于 0 则说明没有访问
+ if (nums[i] > 0)
+ // 将索引转换成元素
+ missing = i + 1;
+ return new int[]{dup, missing};
+ }
+}
+```
\ No newline at end of file
From 3360ffda0fd458460a9c4908a0d7c0c58f539373 Mon Sep 17 00:00:00 2001
From: dragon_li <49151509+1097452462@users.noreply.github.com>
Date: Thu, 12 Nov 2020 20:13:10 +0800
Subject: [PATCH 2/3] =?UTF-8?q?Update=20=E7=BC=BA=E5=A4=B1=E5=92=8C?=
=?UTF-8?q?=E9=87=8D=E5=A4=8D=E7=9A=84=E5=85=83=E7=B4=A0.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
高频面试系列/缺失和重复的元素.md | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/高频面试系列/缺失和重复的元素.md b/高频面试系列/缺失和重复的元素.md
index 52c8d78..9e0b550 100644
--- a/高频面试系列/缺失和重复的元素.md
+++ b/高频面试系列/缺失和重复的元素.md
@@ -1,4 +1,4 @@
-# 如何寻找缺失和重复的元素
+# 如何寻找缺失和重复的元素
@@ -165,4 +165,4 @@ class Solution {
return new int[]{dup, missing};
}
}
-```
\ No newline at end of file
+```
From 1b2ec335ea4b1ae65ba4cb1e92ff2b9e318b1084 Mon Sep 17 00:00:00 2001
From: dragon_li <49151509+1097452462@users.noreply.github.com>
Date: Thu, 12 Nov 2020 20:15:23 +0800
Subject: [PATCH 3/3] =?UTF-8?q?Update=20=E7=BC=BA=E5=A4=B1=E5=92=8C?=
=?UTF-8?q?=E9=87=8D=E5=A4=8D=E7=9A=84=E5=85=83=E7=B4=A0.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
高频面试系列/缺失和重复的元素.md | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/高频面试系列/缺失和重复的元素.md b/高频面试系列/缺失和重复的元素.md
index 9e0b550..eb21f4c 100644
--- a/高频面试系列/缺失和重复的元素.md
+++ b/高频面试系列/缺失和重复的元素.md
@@ -148,7 +148,7 @@ class Solution {
int n = nums.length;
int dup = -1;
for (int i = 0; i < n; i++) {
- // 元素是从 1 开始的
+ // 元素是从 1 开始的
int index = Math.abs(nums[i]) - 1;
// nums[index] 小于 0 则说明重复访问
if (nums[index] < 0)
@@ -160,7 +160,7 @@ class Solution {
for (int i = 0; i < n; i++)
// nums[i] 大于 0 则说明没有访问
if (nums[i] > 0)
- // 将索引转换成元素
+ // 将索引转换成元素
missing = i + 1;
return new int[]{dup, missing};
}