287. 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 repeated number in nums, return this repeated number.
You must solve the problem without modifying the array nums and uses only constant extra space.
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| 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:
- 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#
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| class Solution {
/*
[0,1,2,3,4]
[3,1,3,4,2]
slow: 3 4 2
fast: 3 23 4
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xxx xx
xx xx
xxxxxxxxxxxxxxxxxx x
xx x
xxxx xxx
xxxxxxx
*/
public int findDuplicate(int[] nums) {
int fast = nums[0];
int slow = nums[0];
do {
slow = nums[slow];
fast = nums[nums[fast]];
} while (slow != fast);
slow = nums[0];
while (slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return fast;
}
public int findDuplicate1(int[] nums) {
int duplicate = -1;
for (int i = 0; i < nums.length; i++) {
int curr = Math.abs(nums[i]);
if (nums[curr] < 0) {
duplicate = curr;
}
nums[curr] *= -1;
}
for (int i = 0; i < nums.length; i++) {
nums[i] = Math.abs(nums[i]);
}
return duplicate;
}
}
|