Add solution 1734

leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation.go

@@ -0,0 +1,17 @@
+package leetcode
+
+func decode(encoded []int) []int {
+	n, total, odd := len(encoded), 0, 0
+	for i := 1; i <= n+1; i++ {
+		total ^= i
+	}
+	for i := 1; i < n; i += 2 {
+		odd ^= encoded[i]
+	}
+	perm := make([]int, n+1)
+	perm[0] = total ^ odd
+	for i, v := range encoded {
+		perm[i+1] = perm[i] ^ v
+	}
+	return perm
+}

leetcode/1734.Decode-XORed-Permutation/1734. Decode XORed Permutation_test.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1734 struct {
+	para1734
+	ans1734
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1734 struct {
+	encoded []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1734 struct {
+	one []int
+}
+
+func Test_Problem1734(t *testing.T) {
+
+	qs := []question1734{
+
+		{
+			para1734{[]int{3, 1}},
+			ans1734{[]int{1, 2, 3}},
+		},
+
+		{
+			para1734{[]int{6, 5, 4, 6}},
+			ans1734{[]int{2, 4, 1, 5, 3}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1734------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1734, q.para1734
+		fmt.Printf("【input】:%v       【output】:%v\n", p, decode(p.encoded))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1734.Decode-XORed-Permutation/README.md

@@ -0,0 +1,71 @@
+# [1734. Decode XORed Permutation](https://leetcode.com/problems/decode-xored-permutation/)
+
+
+## 题目
+
+There is an integer array `perm` that is a permutation of the first `n` positive integers, where `n` is always **odd**.
+
+It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = perm[i] XOR perm[i + 1]`. For example, if `perm = [1,3,2]`, then `encoded = [2,1]`.
+
+Given the `encoded` array, return *the original array* `perm`. It is guaranteed that the answer exists and is unique.
+
+**Example 1:**
+
+```
+Input: encoded = [3,1]
+Output: [1,2,3]
+Explanation: If perm = [1,2,3], then encoded = [1 XOR 2,2 XOR 3] = [3,1]
+
+```
+
+**Example 2:**
+
+```
+Input: encoded = [6,5,4,6]
+Output: [2,4,1,5,3]
+
+```
+
+**Constraints:**
+
+- `3 <= n < 10^5`
+- `n` is odd.
+- `encoded.length == n - 1`
+
+## 题目大意
+
+给你一个整数数组 perm ，它是前 n 个正整数的排列，且 n 是个奇数 。它被加密成另一个长度为 n - 1 的整数数组 encoded ，满足 encoded[i] = perm[i] XOR perm[i + 1] 。比方说，如果 perm = [1,3,2] ，那么 encoded = [2,1] 。给你 encoded 数组，请你返回原始数组 perm 。题目保证答案存在且唯一。
+
+## 解题思路
+
+- 这一题与第 136 题和第 137 题思路类似，借用 `x ^ x = 0` 这个性质解题。依题意，原数组 perm 是 n 个正整数，即取值在 `[1,n+1]` 区间内，但是排列顺序未知。可以考虑先将 `[1,n+1]` 区间内的所有数异或得到 total。再将 encoded 数组中奇数下标的元素异或得到 odd：
+
+    $$\begin{aligned}odd &= encoded[1] + encoded[3] + ... + encoded[n-1]\\&= (perm[1] \,\, XOR \,\, perm[2]) + (perm[3] \,\,  XOR  \,\, perm[4]) + ... + (perm[n-1]  \,\, XOR \,\, perm[n])\end{aligned}$$
+
+    total 是 n 个正整数异或全集，odd 是 `n-1` 个正整数异或集。两者异或 `total ^ odd` 得到的值必定是 perm[0]，因为 `x ^ x = 0`，那么重复出现的元素被异或以后消失了。算出 perm[0] 就好办了。
+
+    $$\begin{aligned}encoded[0] &= perm[0] \,\, XOR \,\, perm[1]\\perm[0] \,\, XOR \,\, encoded[0] &= perm[0] \,\, XOR \,\, perm[0] \,\, XOR \,\, perm[1] = perm[1]\\perm[1] \,\, XOR \,\, encoded[1] &= perm[1] \,\, XOR \,\, perm[1] \,\, XOR \,\, perm[2] = perm[2]\\...\\perm[n-1] \,\, XOR \,\, encoded[n-1] &= perm[n-1] \,\, XOR \,\, perm[n-1] \,\, XOR \,\, perm[n] = perm[n]\\\end{aligned}$$
+
+    依次类推，便可以推出原数组 perm 中的所有数。
+
+## 代码
+
+```go
+package leetcode
+
+func decode(encoded []int) []int {
+	n, total, odd := len(encoded), 0, 0
+	for i := 1; i <= n+1; i++ {
+		total ^= i
+	}
+	for i := 1; i < n; i += 2 {
+		odd ^= encoded[i]
+	}
+	perm := make([]int, n+1)
+	perm[0] = total ^ odd
+	for i, v := range encoded {
+		perm[i+1] = perm[i] ^ v
+	}
+	return perm
+}
+```