feat: add solutions to lc problems: No.0506,0507

solution/0500-0599/0506.Relative Ranks/README.md

@@ -65,7 +65,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：排序
+
+我们使用一个数组 $\textit{idx}$ 存储 $0$ 到 $n-1$ 的下标，然后对 $\textit{idx}$ 进行排序，排序规则为：按照 $\textit{score}$ 的值从大到小排序。
+
+然后我们定义一个数组 $\textit{top3} = [\text{Gold Medal}, \text{Silver Medal}, \text{Bronze Medal}]$，遍历 $\textit{idx}$，对于每个下标 $j$，如果 $j$ 小于 $3$，则 $\textit{ans}[j]$ 为 $\textit{top3}[j]$，否则为 $j+1$。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{score}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -77,10 +83,10 @@ class Solution:
         n = len(score)
         idx = list(range(n))
         idx.sort(key=lambda x: -score[x])
-        top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal']
+        top3 = ["Gold Medal", "Silver Medal", "Bronze Medal"]
         ans = [None] * n
-        for i in range(n):
-            ans[idx[i]] = top3[i] if i < 3 else str(i + 1)
+        for i, j in enumerate(idx):
+            ans[j] = top3[i] if i < 3 else str(i + 1)
         return ans
 ```
 
@@ -112,15 +118,16 @@ class Solution {
 public:
     vector<string> findRelativeRanks(vector<int>& score) {
         int n = score.size();
-        vector<pair<int, int>> idx;
-        for (int i = 0; i < n; ++i)
-            idx.push_back(make_pair(score[i], i));
-        sort(idx.begin(), idx.end(),
-            [&](const pair<int, int>& x, const pair<int, int>& y) { return x.first > y.first; });
+        vector<int> idx(n);
+        iota(idx.begin(), idx.end(), 0);
+        sort(idx.begin(), idx.end(), [&score](int a, int b) {
+            return score[a] > score[b];
+        });
         vector<string> ans(n);
         vector<string> top3 = {"Gold Medal", "Silver Medal", "Bronze Medal"};
-        for (int i = 0; i < n; ++i)
-            ans[idx[i].second] = i < 3 ? top3[i] : to_string(i + 1);
+        for (int i = 0; i < n; ++i) {
+            ans[idx[i]] = i < 3 ? top3[i] : to_string(i + 1);
+        }
         return ans;
     }
 };
@@ -151,6 +158,26 @@ func findRelativeRanks(score []int) []string {
 }
 ```
 
+#### TypeScript
+
+```ts
+function findRelativeRanks(score: number[]): string[] {
+    const n = score.length;
+    const idx = Array.from({ length: n }, (_, i) => i);
+    idx.sort((a, b) => score[b] - score[a]);
+    const top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal'];
+    const ans: string[] = Array(n);
+    for (let i = 0; i < n; i++) {
+        if (i < 3) {
+            ans[idx[i]] = top3[i];
+        } else {
+            ans[idx[i]] = (i + 1).toString();
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0500-0599/0506.Relative Ranks/README_EN.md

@@ -64,7 +64,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Sorting
+
+We use an array $\textit{idx}$ to store the indices from $0$ to $n-1$, then sort $\textit{idx}$ based on the values in $\textit{score}$ in descending order.
+
+Next, we define an array $\textit{top3} = [\text{Gold Medal}, \text{Silver Medal}, \text{Bronze Medal}]$. We traverse $\textit{idx}$, and for each index $j$, if $j$ is less than $3$, then $\textit{ans}[j]$ is $\textit{top3}[j]$; otherwise, it is $j+1$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{score}$.
 
 <!-- tabs:start -->
 
@@ -76,10 +82,10 @@ class Solution:
         n = len(score)
         idx = list(range(n))
         idx.sort(key=lambda x: -score[x])
-        top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal']
+        top3 = ["Gold Medal", "Silver Medal", "Bronze Medal"]
         ans = [None] * n
-        for i in range(n):
-            ans[idx[i]] = top3[i] if i < 3 else str(i + 1)
+        for i, j in enumerate(idx):
+            ans[j] = top3[i] if i < 3 else str(i + 1)
         return ans
 ```
 
@@ -111,15 +117,16 @@ class Solution {
 public:
     vector<string> findRelativeRanks(vector<int>& score) {
         int n = score.size();
-        vector<pair<int, int>> idx;
-        for (int i = 0; i < n; ++i)
-            idx.push_back(make_pair(score[i], i));
-        sort(idx.begin(), idx.end(),
-            [&](const pair<int, int>& x, const pair<int, int>& y) { return x.first > y.first; });
+        vector<int> idx(n);
+        iota(idx.begin(), idx.end(), 0);
+        sort(idx.begin(), idx.end(), [&score](int a, int b) {
+            return score[a] > score[b];
+        });
         vector<string> ans(n);
         vector<string> top3 = {"Gold Medal", "Silver Medal", "Bronze Medal"};
-        for (int i = 0; i < n; ++i)
-            ans[idx[i].second] = i < 3 ? top3[i] : to_string(i + 1);
+        for (int i = 0; i < n; ++i) {
+            ans[idx[i]] = i < 3 ? top3[i] : to_string(i + 1);
+        }
         return ans;
     }
 };
@@ -150,6 +157,26 @@ func findRelativeRanks(score []int) []string {
 }
 ```
 
+#### TypeScript
+
+```ts
+function findRelativeRanks(score: number[]): string[] {
+    const n = score.length;
+    const idx = Array.from({ length: n }, (_, i) => i);
+    idx.sort((a, b) => score[b] - score[a]);
+    const top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal'];
+    const ans: string[] = Array(n);
+    for (let i = 0; i < n; i++) {
+        if (i < 3) {
+            ans[idx[i]] = top3[i];
+        } else {
+            ans[idx[i]] = (i + 1).toString();
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0500-0599/0506.Relative Ranks/Solution.cpp

@@ -2,15 +2,16 @@ class Solution {
 public:
     vector<string> findRelativeRanks(vector<int>& score) {
         int n = score.size();
-        vector<pair<int, int>> idx;
-        for (int i = 0; i < n; ++i)
-            idx.push_back(make_pair(score[i], i));
-        sort(idx.begin(), idx.end(),
-            [&](const pair<int, int>& x, const pair<int, int>& y) { return x.first > y.first; });
+        vector<int> idx(n);
+        iota(idx.begin(), idx.end(), 0);
+        sort(idx.begin(), idx.end(), [&score](int a, int b) {
+            return score[a] > score[b];
+        });
         vector<string> ans(n);
         vector<string> top3 = {"Gold Medal", "Silver Medal", "Bronze Medal"};
-        for (int i = 0; i < n; ++i)
-            ans[idx[i].second] = i < 3 ? top3[i] : to_string(i + 1);
+        for (int i = 0; i < n; ++i) {
+            ans[idx[i]] = i < 3 ? top3[i] : to_string(i + 1);
+        }
         return ans;
     }
-};
+};

solution/0500-0599/0506.Relative Ranks/Solution.py

@@ -3,8 +3,8 @@ def findRelativeRanks(self, score: List[int]) -> List[str]:
         n = len(score)
         idx = list(range(n))
         idx.sort(key=lambda x: -score[x])
-        top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal']
+        top3 = ["Gold Medal", "Silver Medal", "Bronze Medal"]
         ans = [None] * n
-        for i in range(n):
-            ans[idx[i]] = top3[i] if i < 3 else str(i + 1)
+        for i, j in enumerate(idx):
+            ans[j] = top3[i] if i < 3 else str(i + 1)
         return ans

solution/0500-0599/0506.Relative Ranks/Solution.ts

@@ -0,0 +1,15 @@
+function findRelativeRanks(score: number[]): string[] {
+    const n = score.length;
+    const idx = Array.from({ length: n }, (_, i) => i);
+    idx.sort((a, b) => score[b] - score[a]);
+    const top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal'];
+    const ans: string[] = Array(n);
+    for (let i = 0; i < n; i++) {
+        if (i < 3) {
+            ans[idx[i]] = top3[i];
+        } else {
+            ans[idx[i]] = (i + 1).toString();
+        }
+    }
+    return ans;
+}

solution/0500-0599/0507.Perfect Number/README.md

@@ -51,7 +51,15 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：枚举
+
+我们首先判断 $\textit{num}$ 是否为 1，如果为 1，则 $\textit{num}$ 不是完美数，返回 $\text{false}$。
+
+然后，我们从 2 开始枚举 $\textit{num}$ 的所有正因子，如果 $\textit{num}$ 能被 $\textit{num}$ 的某个正因子 $i$ 整除，那么我们将 $i$ 加入到答案 $\textit{s}$ 中。如果 $\textit{num}$ 除以 $i$ 得到的商不等于 $i$，我们也将 $\textit{num}$ 除以 $i$ 得到的商加入到答案 $\textit{s}$ 中。
+
+最后，我们判断 $\textit{s}$ 是否等于 $\textit{num}$ 即可。
+
+时间复杂度 $O(\sqrt{n})$，其中 $n$ 为 $\textit{num}$ 的大小。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -63,7 +71,7 @@ class Solution:
         if num == 1:
             return False
         s, i = 1, 2
-        while i * i <= num:
+        while i <= num // i:
             if num % i == 0:
                 s += i
                 if i != num // i:
@@ -76,13 +84,12 @@ class Solution:
 
 ```java
 class Solution {
-
     public boolean checkPerfectNumber(int num) {
         if (num == 1) {
             return false;
         }
         int s = 1;
-        for (int i = 2; i * i <= num; ++i) {
+        for (int i = 2; i <= num / i; ++i) {
             if (num % i == 0) {
                 s += i;
                 if (i != num / i) {
@@ -101,12 +108,16 @@ class Solution {
 class Solution {
 public:
     bool checkPerfectNumber(int num) {
-        if (num == 1) return false;
+        if (num == 1) {
+            return false;
+        }
         int s = 1;
-        for (int i = 2; i * i <= num; ++i) {
+        for (int i = 2; i <= num / i; ++i) {
             if (num % i == 0) {
                 s += i;
-                if (i != num / i) s += num / i;
+                if (i != num / i) {
+                    s += num / i;
+                }
             }
         }
         return s == num;
@@ -122,18 +133,38 @@ func checkPerfectNumber(num int) bool {
 		return false
 	}
 	s := 1
-	for i := 2; i*i <= num; i++ {
+	for i := 2; i <= num/i; i++ {
 		if num%i == 0 {
 			s += i
-			if i != num/i {
-				s += num / i
+			if j := num / i; i != j {
+				s += j
 			}
 		}
 	}
 	return s == num
 }
 ```
 
+#### TypeScript
+
+```ts
+function checkPerfectNumber(num: number): boolean {
+    if (num <= 1) {
+        return false;
+    }
+    let s = 1;
+    for (let i = 2; i <= num / i; ++i) {
+        if (num % i === 0) {
+            s += i;
+            if (i * i !== num) {
+                s += num / i;
+            }
+        }
+    }
+    return s === num;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->