feat: update leetcode solutions: No.0287. Find the Duplicate Number

Author: yanglbme

solution/0200-0299/0287.Find the Duplicate Number/README.md

@@ -32,22 +32,74 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+二分法。
+
+如果值范围在 `[1, mid]` 的数小于等于 mid，说明此范围内没有重复的数，否则说明有重复数。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def findDuplicate(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        while l < r:
+            mid = (l + r) >> 1
+            cnt = 0
+            for e in nums:
+                if e <= mid:
+                    cnt += 1
+            if cnt <= mid:
+                l = mid + 1
+            else:
+                r = mid
+        return l
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int findDuplicate(int[] nums) {
+        int l = 1, r = nums.length - 1;
+        while (l < r) {
+            int mid = (l + r) >>> 1;
+            int cnt = 0;
+            for (int e : nums) {
+                if (e <= mid) ++cnt;
+            }
+            if (cnt <= mid) l = mid + 1;
+            else r = mid;
+        }
+        return l;
+    }
+}
+```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int findDuplicate(vector<int>& nums) {
+        int l = 0, r = nums.size() - 1;
+        while (l < r) {
+            int mid = l + ((r - l) >> 1);
+            int cnt = 0;
+            for (auto e : nums) {
+                if (e <= mid) ++cnt;
+            }
+            if (cnt <= mid) l = mid + 1;
+            else r = mid;
+        }
+        return l;
+    }
+};
 ```
 
 ### **...**

solution/0200-0299/0287.Find the Duplicate Number/README_EN.md

@@ -40,13 +40,61 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def findDuplicate(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        while l < r:
+            mid = (l + r) >> 1
+            cnt = 0
+            for e in nums:
+                if e <= mid:
+                    cnt += 1
+            if cnt <= mid:
+                l = mid + 1
+            else:
+                r = mid
+        return l
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int findDuplicate(int[] nums) {
+        int l = 1, r = nums.length - 1;
+        while (l < r) {
+            int mid = (l + r) >>> 1;
+            int cnt = 0;
+            for (int e : nums) {
+                if (e <= mid) ++cnt;
+            }
+            if (cnt <= mid) l = mid + 1;
+            else r = mid;
+        }
+        return l;
+    }
+}
+```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int findDuplicate(vector<int>& nums) {
+        int l = 0, r = nums.size() - 1;
+        while (l < r) {
+            int mid = l + ((r - l) >> 1);
+            int cnt = 0;
+            for (auto e : nums) {
+                if (e <= mid) ++cnt;
+            }
+            if (cnt <= mid) l = mid + 1;
+            else r = mid;
+        }
+        return l;
+    }
+};
 ```
 
 ### **...**

solution/0200-0299/0287.Find the Duplicate Number/Solution.cpp

@@ -0,0 +1,16 @@
+class Solution {
+public:
+    int findDuplicate(vector<int>& nums) {
+        int l = 0, r = nums.size() - 1;
+        while (l < r) {
+            int mid = l + ((r - l) >> 1);
+            int cnt = 0;
+            for (auto e : nums) {
+                if (e <= mid) ++cnt;
+            }
+            if (cnt <= mid) l = mid + 1;
+            else r = mid;
+        }
+        return l;
+    }
+};

solution/0200-0299/0287.Find the Duplicate Number/Solution.java

@@ -1,19 +1,15 @@
-public class Solution {
-    // https://segmentfault.com/a/1190000003817671
+class Solution {
     public int findDuplicate(int[] nums) {
-        int slow = 0;
-        int fast = 0;
-        // 找到快慢指针相遇的地方
-        do{
-            slow = nums[slow];
-            fast = nums[nums[fast]];
-        } while(slow != fast);
-        int find = 0;
-        // 用一个新指针从头开始，直到和慢指针相遇
-        while(find != slow){
-            slow = nums[slow];
-            find = nums[find];
+        int l = 1, r = nums.length - 1;
+        while (l < r) {
+            int mid = (l + r) >>> 1;
+            int cnt = 0;
+            for (int e : nums) {
+                if (e <= mid) ++cnt;
+            }
+            if (cnt <= mid) l = mid + 1;
+            else r = mid;
         }
-        return find;
+        return l;
     }
 }

solution/0200-0299/0287.Find the Duplicate Number/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def findDuplicate(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        while l < r:
+            mid = (l + r) >> 1
+            cnt = 0
+            for e in nums:
+                if e <= mid:
+                    cnt += 1
+            if cnt <= mid:
+                l = mid + 1
+            else:
+                r = mid
+        return l