diff --git "a/Index/\346\250\241\346\213\237.md" "b/Index/\346\250\241\346\213\237.md"
index 32d75701..bcfb45cf 100644
--- "a/Index/\346\250\241\346\213\237.md"
+++ "b/Index/\346\250\241\346\213\237.md"
@@ -64,6 +64,7 @@
 | [645. 错误的集合](https://leetcode-cn.com/problems/set-mismatch/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/set-mismatch/solution/gong-shui-san-xie-yi-ti-san-jie-ji-shu-s-vnr9/) | 简单 | 🤩🤩🤩 |
 | [709. 转换成小写字母](https://leetcode-cn.com/problems/to-lower-case/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/to-lower-case/solution/gong-shui-san-xie-jian-dan-zi-fu-chuan-m-czpo/) | 简单 | 🤩🤩🤩 |
 | [726. 原子的数量](https://leetcode-cn.com/problems/number-of-atoms/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/number-of-atoms/solution/gong-shui-san-xie-shi-yong-xiao-ji-qiao-l5ak4/) | 困难 | 🤩🤩🤩🤩 |
+| [747. 至少是其他数字两倍的最大数](https://leetcode-cn.com/problems/largest-number-at-least-twice-of-others/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/largest-number-at-least-twice-of-others/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-8179/) | 简单 | 🤩🤩🤩🤩🤩 |
 | [748. 最短补全词](https://leetcode-cn.com/problems/shortest-completing-word/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/shortest-completing-word/solution/gong-shui-san-xie-jian-dan-zi-fu-chuan-j-x4ao/) | 简单 | 🤩🤩🤩🤩 |
 | [766. 托普利茨矩阵](https://leetcode-cn.com/problems/toeplitz-matrix/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/toeplitz-matrix/solution/cong-ci-pan-du-qu-cheng-ben-fen-xi-liang-f20w/) | 简单 | 🤩🤩🤩 |
 | [794. 有效的井字游戏](https://leetcode-cn.com/problems/valid-tic-tac-toe-state/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/valid-tic-tac-toe-state/solution/gong-shui-san-xie-fen-qing-kuang-tao-lun-pikn/) | 中等 | 🤩🤩🤩🤩 |
diff --git "a/LeetCode/741-750/747. \350\207\263\345\260\221\346\230\257\345\205\266\344\273\226\346\225\260\345\255\227\344\270\244\345\200\215\347\232\204\346\234\200\345\244\247\346\225\260\357\274\210\347\256\200\345\215\225\357\274\211.md" "b/LeetCode/741-750/747. \350\207\263\345\260\221\346\230\257\345\205\266\344\273\226\346\225\260\345\255\227\344\270\244\345\200\215\347\232\204\346\234\200\345\244\247\346\225\260\357\274\210\347\256\200\345\215\225\357\274\211.md"
new file mode 100644
index 00000000..9e811262
--- /dev/null
+++ "b/LeetCode/741-750/747. \350\207\263\345\260\221\346\230\257\345\205\266\344\273\226\346\225\260\345\255\227\344\270\244\345\200\215\347\232\204\346\234\200\345\244\247\346\225\260\357\274\210\347\256\200\345\215\225\357\274\211.md"	
@@ -0,0 +1,408 @@
+### 题目描述
+
+这是 LeetCode 上的 **[747. 至少是其他数字两倍的最大数](https://leetcode-cn.com/problems/largest-number-at-least-twice-of-others/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-8179/)** ，难度为 **简单**。
+
+Tag : 「模拟」
+
+
+
+给你一个整数数组 $nums$ ，其中总是存在 **唯一的** 一个最大整数 。
+
+请你找出数组中的最大元素并检查它是否 至少是数组中每个其他数字的两倍 。
+
+如果是，则返回 **最大元素的下标** ，否则返回 $-1$ 。
+
+示例 1：
+```
+输入：nums = [3,6,1,0]
+
+输出：1
+
+解释：6 是最大的整数，对于数组中的其他整数，6 大于数组中其他元素的两倍。6 的下标是 1 ，所以返回 1 。
+```
+
+示例 2：
+```
+输入：nums = [1,2,3,4]
+输出：-1
+解释：4 没有超过 3 的两倍大，所以返回 -1 。
+```
+示例 3：
+```
+输入：nums = [1]
+
+输出：0
+
+解释：因为不存在其他数字，所以认为现有数字 1 至少是其他数字的两倍。
+```
+
+提示：
+* $1 <= nums.length <= 50$
+* $0 <= nums[i] <= 100$
+* $nums$ 中的最大元素是唯一的
+
+---
+
+### 模拟
+
+根据题意进行模拟即可，遍历过程中维护最大值和次大值的下标，最后判断最大值是否至少为次大值两倍。
+
+然后？今天属于圆梦了？（这真的只是他们的日常可爱 🤣
+
+![image.png](https://pic.leetcode-cn.com/1642041486-TCwCib-image.png)
+
+**代码（感谢 [@5cm/s 🌸](/u/megurine/)、[@Benhao](/u/himymben/) 和 [@Qian](/u/qian2/) 几位总提供的其他语言版本 🤣 🤣 ）：**
+```Java
+class Solution {
+    public int dominantIndex(int[] nums) {
+        int n = nums.length;
+        if (n == 1) return 0;
+        int a = -1, b = 0;
+        for (int i = 1; i < n; i++) {
+            if (nums[i] > nums[b]) {
+                a = b; b = i;
+            } else if (a == -1 || nums[i] > nums[a]) {
+                a = i;
+            }
+        }
+        return nums[b] >= nums[a] * 2 ? b : -1;
+    }
+}
+```
+-
+```Python
+class Solution(object):
+    def dominantIndex(self, nums):
+        n = len(nums)
+        if n == 1:
+            return 0
+        a, b = -1, 0
+        for i in range(1, n):
+            if nums[i] > nums[b]:
+                a, b = b, i
+            elif a == -1 or nums[i] > nums[a]:
+                a = i
+        return b if nums[b] >= nums[a] * 2 else -1
+```
+-
+```Python
+class Solution:
+    def dominantIndex(self, nums: List[int]) -> int:
+        n = len(nums)
+        if n == 1:
+            return 0
+        a, b = -1, 0
+        for i in range(1, n):
+            if nums[i] > nums[b]:
+                a, b = b, i
+            elif a == -1 or nums[i] > nums[a]:
+                a = i
+        return b if nums[b] >= nums[a] * 2 else -1
+```
+-
+```Go
+func dominantIndex(nums []int) int {
+    n := len(nums)
+    if n == 1{
+        return 0
+    }
+    a, b := -1, 0
+    for i := 1; i < n; i++ {
+        if nums[i] > nums[b] {
+            a, b = b, i
+        } else if a == -1 || nums[i] > nums[a] {
+            a = i
+        }
+    }
+    if nums[b] >= nums[a] * 2{
+        return b
+    }
+    return -1
+}
+```
+-
+```Rust
+impl Solution {
+    pub fn dominant_index(nums: Vec<i32>) -> i32 {
+        let mut mx = 0;
+        let mut ans = -1 as i32;
+        for i in 0..nums.len() {
+            if nums[i] >= mx * 2 {
+                ans = i as i32;
+            }
+            else if nums[i] * 2 > mx {
+                ans = -1 as i32;
+            }
+            if nums[i] > mx {
+                mx = nums[i];
+            }
+        }
+        ans
+    }
+}
+```
+-
+```C
+int dominantIndex(int* nums, int numsSize){
+    if(numsSize == 1) return 0;
+    int a = -1, b = 0;
+    for(int i = 1; i < numsSize; i++){
+        if (nums[i] > nums[b]) {
+            a = b; b = i;
+        } else if (a == -1 || nums[i] > nums[a]) {
+            a = i;
+        }
+    }
+    return nums[b] >= nums[a] * 2 ? b : -1;
+}
+```
+-
+```C#
+public class Solution {
+    public int DominantIndex(int[] nums) {
+        int n = nums.Length;
+        if (n == 1) return 0;
+        int a = -1, b = 0;
+        for (int i = 1; i < n; i++) {
+            if (nums[i] > nums[b]) {
+                a = b; b = i;
+            } else if (a == -1 || nums[i] > nums[a]) {
+                a = i;
+            }
+        }
+        return nums[b] >= nums[a] * 2 ? b : -1;
+    }
+}
+```
+-
+```C++
+class Solution {
+public:
+    int dominantIndex(vector<int>& nums) {
+        int mx = 0, ans = -1;
+        for (int i = 0; i < nums.size(); ++i) {
+            if (nums[i] >= mx * 2) ans = i;
+            else if (nums[i] * 2 > mx) ans = -1;
+            mx = max(nums[i], mx);
+        }
+        return ans;
+    }
+};
+```
+-
+```JS
+var dominantIndex = function(nums) {
+    const n = nums.length
+    if(n == 1) return 0
+    let a = -1, b = 0
+    for (let i = 1; i < n; i++){
+        if (nums[i] > nums[b]) {
+            a = b; b = i;
+        } else if (a == -1 || nums[i] > nums[a]) {
+            a = i;
+        }
+    }
+    return nums[b] >= nums[a] * 2 ? b : -1;
+};
+```
+-
+```TS
+function dominantIndex(nums: number[]): number {
+    const n = nums.length
+    if(n == 1) return 0
+    let a = -1, b = 0
+    for (let i = 1; i < n; i++){
+        if (nums[i] > nums[b]) {
+            a = b; b = i;
+        } else if (a == -1 || nums[i] > nums[a]) {
+            a = i;
+        }
+    }
+    return nums[b] >= nums[a] * 2 ? b : -1;
+};
+```
+-
+```PHP
+class Solution {
+    function dominantIndex($nums) {
+        $n = sizeof($nums);
+        if ($n == 1) return 0;
+        $a = -1; $b = 0;
+        for ($i = 1; $i < $n; $i++) {
+            if ($nums[$i] > $nums[$b]) {
+                $a = $b;
+                $b = $i;
+            } else if ($a == -1 || $nums[$i] > $nums[$a]) {
+                $a = $i;
+            }
+        }
+        if($nums[$b] >= $nums[$a] * 2){
+            return $b;
+        }
+        return -1;
+    }
+}
+```
+-
+```Swift
+class Solution {
+    func dominantIndex(_ nums: [Int]) -> Int {
+        var n = nums.count
+        if n == 1 {
+            return 0
+        }
+        var (a, b) = (-1, 0)
+        for i in 1..<nums.count {
+            if nums[i] > nums[b] {
+                (a, b) = (b, i)
+            }
+            else if a == -1 || nums[i] > nums[a] {
+                a = i
+            }
+        }
+        return nums[b] >= nums[a] * 2 ? b : -1
+    }
+}
+```
+-
+```Kotlin
+class Solution {
+    fun dominantIndex(nums: IntArray): Int {
+        var mx = 0
+        var ans = -1
+        for (i in nums.indices) {
+            when {
+                nums[i] >= mx * 2 -> ans = i
+                nums[i] * 2 > mx -> ans = -1
+            }
+            if (nums[i] > mx) mx = nums[i]
+        }
+        return ans
+    }
+}
+```
+-
+```Scala
+object Solution {
+    def dominantIndex(nums: Array[Int]): Int = {
+        var mx = 0
+        var ans = -1
+        for (i <- 0 to (nums.length - 1)) {
+            if (nums(i) >= mx * 2) ans = i
+            else if (nums(i) * 2 > mx) ans = -1
+            if (nums(i) > mx) mx = nums(i)
+        }
+        return ans
+    }
+}
+```
+-
+```Ruby
+def dominant_index(nums)
+    n = nums.length
+    return 0 if n == 1
+    a = -1
+    b = 0
+    nums.each_with_index do |num, i|
+        next if i == 0
+        if nums[i] > nums[b]
+            a = b
+            b = i
+        elsif a.eql?(-1) or nums[i] > nums[a]
+            a = i
+        end
+    end
+    return nums[b] >= nums[a] * 2 ? b : -1
+end
+```
+-
+```lisp
+(define/contract (dominant-index nums)
+  (-> (listof exact-integer?) exact-integer?)
+  (let loop ([nums nums] [i 0] [mx 0] [ans -1])
+    (cond
+      [(empty? nums) ans]
+      [else
+        (define x (car nums))
+        (define mxx (max x mx))
+        (cond
+          [(>= x (* mx 2)) (loop (cdr nums) (+ i 1) mxx i)]
+          [(> (* x 2) mx) (loop (cdr nums) (+ i 1) mxx -1)]
+          [else (loop (cdr nums) (+ i 1) mxx ans)]
+        )
+      ]
+    )
+  )
+)
+```
+-
+
+```Elixir
+defmodule Solution do
+  @spec dominant_index(nums :: [integer]) :: integer
+  def dominant_index(nums) do
+    solve(nums, 0, -1, 0)
+  end
+  def solve([], mx, ans, cur) do
+    ans
+  end
+  def solve([x | rest], mx, ans, cur) when x >= mx * 2 do
+    solve(rest, max(x, mx), cur, cur + 1)
+  end
+  def solve([x | rest], mx, ans, cur) when x * 2 > mx do
+    solve(rest, max(x, mx), -1, cur + 1)
+  end
+  def solve([x | rest], mx, ans, cur) do
+    solve(rest, max(x, mx), ans, cur + 1)
+  end
+end
+```
+-
+```Erlang
+-spec dominant_index(Nums :: [integer()]) -> integer().
+dominant_index(Nums) ->
+  solve(Nums).
+
+solve(List) ->
+  solve(List, 0, -1, 0).
+solve([], Max, Ans, Cur) ->
+  Ans;
+solve([Head|Rest], Max, Ans, Cur) when Head >= Max * 2 ->
+  solve(Rest, max(Max, Head), Cur, Cur + 1);
+solve([Head|Rest], Max, Ans, Cur) when Head * 2 > Max ->
+  solve(Rest, max(Max, Head), -1, Cur + 1);
+solve([Head|Rest], Max, Ans, Cur) ->
+  solve(Rest, max(Max, Head), Ans, Cur + 1).
+
+max(A, B) when A > B ->
+  A;
+max(A, B) ->
+  B.
+```
+* 时间复杂度：$O(n)$
+* 空间复杂度：$O(1)$
+
+---
+
+### 全鱼宴（是全语言 这错别字还挺可爱 🤣
+
+经过三位总 [@5cm/s 🌸](/u/megurine/)、[@Benhao](/u/himymben/) 和 [@Qian](/u/qian2/) 的不懈努力，通过举一反三、连蒙带猜，把所有语言弄出了 🤣
+
+这个过程中也有幸见到，如果一个语言只有一份代码，居然界面是这样的（没有时间和内存的分布图：
+
+![image.png](https://pic.leetcode-cn.com/1642054606-AiJDCs-image.png)
+
+我和我的小伙伴纷纷表示这样的 Flag 再也不敢了 🤣
+
+---
+
+### 最后
+
+这是我们「刷穿 LeetCode」系列文章的第 `No.747` 篇，系列开始于 2021/01/01，截止于起始日 LeetCode 上共有 1916 道题目，部分是有锁题，我们将先把所有不带锁的题目刷完。
+
+在这个系列文章里面，除了讲解解题思路以外，还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。
+
+为了方便各位同学能够电脑上进行调试和提交代码，我建立了相关的仓库：https://github.com/SharingSource/LogicStack-LeetCode 。
+
+在仓库地址里，你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。
+