From 96900391887d344a0642bd6623f63e8aee9ce5ec Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Thu, 30 May 2024 15:52:24 +0800
Subject: [PATCH] leetcode udpate

---
 content/posts/leetcode.md | 44 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index ebb85d1..f5b6e52 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -6603,3 +6603,47 @@ func numSteps(s string) int {
 ### 总结
 
 本题直接从最后一位开始向前遍历, 遇到0直接将结果加1, 遇到1设定一个标记进位的标记变量为1, 同时将结果加2, 如此重复即可. 注意在每次判断当前位是0还是1之前要将原本的当前位值加上进位的标记变量再判断.
+
+## day94 2024-05-30
+### 1442. Count Triplets That Can Form Two Arrays of Equal XOR 
+Given an array of integers arr.
+
+We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).
+
+Let's define a and b as follows:
+
+a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
+b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
+Note that ^ denotes the bitwise-xor operation.
+
+Return the number of triplets (i, j and k) Where a == b.
+
+![0530uVcNw3JX4b49](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0530uVcNw3JX4b49.png)  
+
+### 题解
+
+题目中要求i-j和j-k之间的数组元素全部异或后的值相等, 而i<j<=k. 两个相同的值异或后的值为0, 因此可以得出结论, i-k之间的数异或后的值为0(i-j,j-k范围内的数异或后的值相同). i-k区间内的数异或后的值为0, 就意味着从头开始到i和从头开始到k这两个区间的数异或的值相等(不一定为0). 遍历数组, 计算到当前下标处的异或值. 并将异或值作为key, 下标作为value保存到map中, 计算得到异或值后查询map, 找到与当前下标异或值相等的全部下标, 并将这些区间内的所有可行分割(长度为n的区间有n-1种分割方法, 直观理解放木板到这个区间的数字之间, 一共能放几块就是几种分割方法)数目添加到结果中.  
+
+### 代码
+```go 
+func countTriplets(arr []int) int {
+    xormap := map[int][]int{}
+    xormap[0] = []int{-1}
+    xornow := 0
+    result := 0
+    beforeslice := []int{}
+    for index, value := range arr{
+        xornow = xornow ^ value
+        beforeslice = xormap[xornow]
+        if len(beforeslice) == 0{
+            xormap[xornow] = []int{index}
+        }else{
+            for _, sameindex := range beforeslice{
+                result += index - sameindex - 1
+            }
+            xormap[xornow] = append(xormap[xornow], index)
+        }
+    }
+    return result
+}
+```