LeetCode - Medium - 287. Find the Duplicate Number

柔光的暖阳◎ 2022-12-29 04:42 135阅读 0赞

## Topic ##

*  Array
 *  Two Pointers
 *  Binary Search

## Description ##

[https://leetcode.com/problems/find-the-duplicate-number/][https_leetcode.com_problems_find-the-duplicate-number]

Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.

There is only `one duplicate number` in nums, return this duplicate number.

**Follow-ups**:

*  How can we prove that at least one duplicate number must exist in `nums`?
 *  Can you solve the problem **without** modifying the array `nums`?
 *  Can you solve the problem using only constant, `O(1)` extra space?
 *  Can you solve the problem with runtime complexity less than `O(n²)`?

**Example 1**:

Input: nums = [1,3,4,2,2]
    Output: 2

**Example 2**:

Input: nums = [3,1,3,4,2]
    Output: 3

**Example 3**:

Input: nums = [1,1]
    Output: 1

**Example 4**:

Input: nums = [1,1,2]
    Output: 1

**Constraints**:

*  2 <= n <= 3 \* 10⁴
 *  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.

## Analysis ##

方法一：归位法，遍历数组，将数值转移到与下标相等的位置，若发现该下标位置的数已有与下标相等的数，则返回该下标值。该法的**副作用**是修改了数组内容。

--------------------

方法二：将该问题转换成[Linked List Cycle II][]。在脑里，将数组转换成链表形式，也就是从下标0开始，下标为0的元素则是链表的头节点的值，接着头节点的值为下标，从数组得出下一节点的值，以此类推，最后会得出带有环的链表。这样方便理解，如符合题意要求的数组：

转换成的链表：

┌─────────┐
         ↓         |
    1 -> 3 -> 2 -> 4

链表中环的入口节点的数值则是题目要求的重复值。接着快慢双指针求出这个值。

--------------------

方法三：二分查找

At first the search space is numbers between 1 to n. Each time I select a number `mid` (which is the one in the middle) and `count` all the numbers equal to or less than `mid`. Then if the count is more than mid, the search space will be `[1, mid]` otherwise `[mid+1, n]`. I do this until search space is only one number.

Let’s say `n=10` and I select `mid=5`. Then I count all the numbers in the array which are **less than equal (<=)** mid. If the there are more than 5 numbers and other are less than 5, then by Pigeonhole Principle (https://en.wikipedia.org/wiki/Pigeonhole\_principle) one of them has occurred more than once. So I shrink the search space from \[1,10\] to \[1,5\] (count >= mid, the duplicate number is in \[1,5\]). Otherwise the duplicate number is in the second half so for the next step the search space would be \[6, 10\].

## Submission ##

public class FindTheDuplicateNumber { 
    
    	// 方法一：
    	public int findDuplicate1(int[] nums) { 
    
    		for (int i = 0; i < nums.length; i++) { 
    			int should = i + 1;
    
    			while (nums[i] != should) { 
    				if (nums[i] == nums[nums[i] - 1]) { 
    					return nums[i];
    				}
    
    				swap(nums, i, nums[i] - 1);
    			}
    		}
    
    		return -1;
    	}
    
    	private void swap(int[] nums, int index1, int index2) { 
    		int temp = nums[index1];
    		nums[index1] = nums[index2];
    		nums[index2] = temp;
    	}
    
    	// 方法二
    	public int findDuplicate2(int[] nums) { 
    
    		if (nums.length > 1) { 
    
    			int slow = nums[0];
    			int fast = nums[nums[0]];
    
    			while (slow != fast) { 
    				slow = nums[slow];
    				fast = nums[nums[fast]];
    			}
    
    			fast = 0;
    			while (fast != slow) { 
    				fast = nums[fast];
    				slow = nums[slow];
    			}
    			return slow;
    		}
    		return -1;
    	}
    
    	// 方法三
    	public int findDuplicate3(int[] nums) { 
    		int low = 1, high = nums.length - 1;
    		while (low < high) { 
    			int mid = low + (high - low) / 2;
    			int cnt = 0;
    			for (int a : nums) { 
    				if (a <= mid)
    					++cnt;
    			}
    			if (cnt <= mid)
    				low = mid + 1;
    			else
    				high = mid;
    		}
    		return low;
    	}
    
    }

## Test ##

import static org.junit.Assert.*;
    import org.junit.Test;
    
    public class FindTheDuplicateNumberTest { 
    
    	@Test
    	public void test() { 
    		FindTheDuplicateNumber obj = new FindTheDuplicateNumber();
    
    		assertEquals(2, obj.findDuplicate1(new int[] { 1, 3, 4, 2, 2}));
    		assertEquals(3, obj.findDuplicate1(new int[] { 3, 1, 3, 4, 2}));
    		assertEquals(1, obj.findDuplicate1(new int[] { 1, 1}));
    		assertEquals(1, obj.findDuplicate1(new int[] { 1, 1, 2}));
    		
    		assertEquals(2, obj.findDuplicate2(new int[] { 1, 3, 4, 2, 2}));
    		assertEquals(3, obj.findDuplicate2(new int[] { 3, 1, 3, 4, 2}));
    		assertEquals(1, obj.findDuplicate2(new int[] { 1, 1}));
    		assertEquals(1, obj.findDuplicate2(new int[] { 1, 1, 2}));
    		
    		assertEquals(2, obj.findDuplicate3(new int[] { 1, 3, 4, 2, 2}));
    		assertEquals(3, obj.findDuplicate3(new int[] { 3, 1, 3, 4, 2}));
    		assertEquals(1, obj.findDuplicate3(new int[] { 1, 1}));
    		assertEquals(1, obj.findDuplicate3(new int[] { 1, 1, 2}));
    	}
    }

[https_leetcode.com_problems_find-the-duplicate-number]: https://leetcode.com/problems/find-the-duplicate-number/
[Linked List Cycle II]: https://leetcode.com/problems/linked-list-cycle-ii/