feat: update solutions to lc problems: No.0503,0504

* No.0503.Next Greater Element II
* No.0504.Base 7

Author: yanglbme

solution/0500-0599/0503.Next Greater Element II/README.md

@@ -255,18 +255,16 @@ func nextGreaterElements(nums []int) []int {
 
 ```ts
 function nextGreaterElements(nums: number[]): number[] {
-    let stack: number[] = [],
+    const stack: number[] = [],
         len = nums.length;
-    let res: number[] = new Array(len).fill(-1);
+    const res: number[] = new Array(len).fill(-1);
     for (let i = 0; i < 2 * len - 1; i++) {
-        while (
-            stack.length !== 0 &&
-            nums[stack[stack.length - 1]] < nums[i % len]
-        ) {
-            res[stack[stack.length - 1]] = nums[i % len];
+        const j = i % len;
+        while (stack.length !== 0 && nums[stack[stack.length - 1]] < nums[j]) {
+            res[stack[stack.length - 1]] = nums[j];
             stack.pop();
         }
-        stack.push(i % len);
+        stack.push(j);
     }
     return res;
 }

solution/0500-0599/0503.Next Greater Element II/README_EN.md

@@ -243,18 +243,16 @@ func nextGreaterElements(nums []int) []int {
 
 ```ts
 function nextGreaterElements(nums: number[]): number[] {
-    let stack: number[] = [],
+    const stack: number[] = [],
         len = nums.length;
-    let res: number[] = new Array(len).fill(-1);
+    const res: number[] = new Array(len).fill(-1);
     for (let i = 0; i < 2 * len - 1; i++) {
-        while (
-            stack.length !== 0 &&
-            nums[stack[stack.length - 1]] < nums[i % len]
-        ) {
-            res[stack[stack.length - 1]] = nums[i % len];
+        const j = i % len;
+        while (stack.length !== 0 && nums[stack[stack.length - 1]] < nums[j]) {
+            res[stack[stack.length - 1]] = nums[j];
             stack.pop();
         }
-        stack.push(i % len);
+        stack.push(j);
     }
     return res;
 }

solution/0500-0599/0503.Next Greater Element II/Solution.ts

@@ -1,16 +1,14 @@
 function nextGreaterElements(nums: number[]): number[] {
-    let stack: number[] = [],
+    const stack: number[] = [],
         len = nums.length;
-    let res: number[] = new Array(len).fill(-1);
+    const res: number[] = new Array(len).fill(-1);
     for (let i = 0; i < 2 * len - 1; i++) {
-        while (
-            stack.length !== 0 &&
-            nums[stack[stack.length - 1]] < nums[i % len]
-        ) {
-            res[stack[stack.length - 1]] = nums[i % len];
+        const j = i % len;
+        while (stack.length !== 0 && nums[stack[stack.length - 1]] < nums[j]) {
+            res[stack[stack.length - 1]] = nums[j];
             stack.pop();
         }
-        stack.push(i % len);
+        stack.push(j);
     }
     return res;
 }

solution/0500-0599/0504.Base 7/README.md

@@ -36,6 +36,12 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：模拟**
+
+我们不妨假设 `num` 大于等于 $0$，那么，如果 `num` 等于 $0$，只需要返回 $0$ 即可。否则，我们将 $num$ 模 $7$ 的结果保存起来，最后逆序拼接成字符串即可。
+
+时间复杂度 $O(\log n)$，忽略答案的空间消耗，空间复杂度 $O(1)$。其中 $n$ 是 `num` 的绝对值大小。
+
 <!-- tabs:start -->
 
 ### **Python3**