From a5552c4d124b0c6d4230a033a6466ce05e42cbba Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Sat, 2 Mar 2024 13:14:22 +0800
Subject: [PATCH] add leetcode content

---
 content/posts/leetcode.md | 76 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 76 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 480de0a..af4c4f9 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -265,3 +265,79 @@ func maximumOddBinaryNumber(s string) string {
 ### 总结
 
 在处理字符串的时候像中间某个位置插入字符也要使用双引号, 如插入字符0要用`+"0"`而不是`+'0'`, 此外在截取切片的时候go的切片是左闭右开的. 如[0:3]截取的是0,1,2三个数
+
+## day5 2024-03-02
+
+### 977. Squares of a Sorted Array
+
+Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.
+
+Example 1:
+
+Input: nums = [-4,-1,0,3,10]
+Output: [0,1,9,16,100]
+Explanation: After squaring, the array becomes [16,1,0,9,100].
+After sorting, it becomes [0,1,9,16,100].
+
+Example 2:
+
+Input: nums = [-7,-3,2,3,11]
+Output: [4,9,9,49,121]
+
+### 题解
+
+考虑原数组已经按照非递减排序的情况下, 找到数组中正负交界处元素, 即数组中第一个正数, 以该位置作为起始位置, 使用双指针法, 分别向前和向后遍历数组, 遍历时不断比较两个指针指向的数字的绝对值大小, 将绝对值小的数字平方后追加到结果数组的尾部, 遍历完成即可完成平方值排序. 这样只需要遍历一遍数组即可完成排序.
+
+### 代码
+
+```go
+func sortedSquares(nums []int) []int {
+    index := 0
+    value := nums[0]
+    var result []int
+    for index, value = range nums{
+        if(value >= 0){
+            break
+        }
+    }
+    backward := index - 1
+    forward := index
+    if index != 0{
+        for _,_ = range nums{
+            if backward<0{
+                result = append(result, nums[forward]*nums[forward])
+                forward++
+            } else if forward>= len(nums){
+                result = append(result, nums[backward] * nums[backward])
+                backward--
+            } else{
+            if(abs(nums[forward]) < abs(nums[backward])){
+                result = append(result, nums[forward]*nums[forward])
+                forward++
+            } else{
+                result = append(result, nums[backward] * nums[backward])
+                backward--
+            }
+            }
+        }
+    }else{
+        for _,_ = range nums{
+            result = append(result, nums[forward]*nums[forward])
+                forward++
+        }
+    }
+    return result
+}
+
+func abs(val int) int {
+    if(val < 0){
+        return -val
+    }else{
+        return val
+    }
+}
+```
+
+### 总结
+
+注意一个go的语法问题, 如果在for range循环中两个变量都使用匿名变量, 则应该使用赋值运算符而不是创建并赋值运算符, 即`for _,_ = range slice` 而不是`for _,_ := range slice`. 这很可能是因为匿名变量默认为已经创建好的变量, 不需要再创建匿名变量本身了.