3.1 Find the Duplicate Number

Description

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note: You must not modify the array (assume the array is read only). You must use only constant, O(1) extra space. Your runtime complexity should be less than O(n2). There is only one duplicate number in the array, but it could be repeated more than once.

Method

Bucket Sort not o (1) space

fast / slow pointers similar with find cycle of linkedlist

Time and Space Complexity

o(n)

Code

not o(1) extra space public class Solution {

public int findDuplicate(int[] nums) {
       if (nums == null || nums.length == 0){
           return 0;
       }
       int n = nums.length;
       int[] arr = new int[n - 1];
       for (int i = 0; i < nums.length; i++){
           if (arr[nums[i] - 1] != 0){
               return nums[i];
           } else {
               arr[nums[i] - 1] = nums[i];
           }
       }
       return 0;
}

}

public class Solution {

public int findDuplicate(int[] nums) {
    if (nums == null || nums.length == 0){
           return 0;
    }
    int slow = 0;
    int fast = 0;
    int finder = 0;

    while (true){

         slow = nums[slow];
         fast = nums[nums[fast]];

         if (slow == fast){
             break;
         }
    }
    while (true){
         slow = nums[slow];
         finder = nums[finder];
         if (slow == finder){
             break;
         }
    }
    return slow;
}

}