Topic
- Array
- Two Pointers
- Binary Search
Description
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：
| 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 3 | 4 | 2 | 3 |
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}));
}
}
来源：oschina
链接：https://my.oschina.net/jallenkwong/blog/4816664