feat: solve No.287

Author: BaffinLee

201-300/287. Find the Duplicate Number.md

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+# 287. Find the Duplicate Number
+
+- Difficulty: Medium.
+- Related Topics: Array, Two Pointers, Binary Search, Bit Manipulation.
+- Similar Questions: First Missing Positive, Single Number, Linked List Cycle II, Missing Number, Set Mismatch.
+
+## Problem
+
+Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
+
+There is only **one repeated number** in `nums`, return **this repeated number**.
+
+You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
+
+ 
+Example 1:
+
+```
+Input: nums = [1,3,4,2,2]
+Output: 2
+```
+
+Example 2:
+
+```
+Input: nums = [3,1,3,4,2]
+Output: 3
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= n <= 105`
+	
+- `nums.length == n + 1`
+	
+- `1 <= nums[i] <= n`
+	
+- All the integers in `nums` appear only **once** except for **precisely one integer** which appears **two or more** times.
+
+
+ 
+**Follow up:**
+
+
+	
+- How can we prove that at least one duplicate number must exist in `nums`?
+	
+- Can you solve the problem in linear runtime complexity?
+
+
+
+## Solution 1
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var findDuplicate = function(nums) {
+  var left = 0;
+  var right = nums.length - 1;
+  while (left < right) {
+      var mid = left + Math.floor((right - left) / 2);
+      var num = getNum(nums, mid);
+      if (num <= mid) {
+          left = mid + 1;
+      } else {
+          right = mid;
+      }
+  }
+  return left;
+};
+
+var getNum = function(nums, n) {
+    var num = 0;
+    for (var i = 0; i < nums.length; i++) {
+        if (nums[i] <= n) num++;
+    }
+    return num;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(1).