Find the repeated item in an array

最新推荐文章于 2022-06-11 14:21:08 发布

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本文介绍两种查找包含从1到n-1范围内整数且恰好有一个重复整数的数组中重复值的方法。一种方法使用O(n)的时间复杂度且不需要额外空间；另一种同样具备O(n)的时间复杂度但需要额外的空间。

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Let A be an array of size n >=2 containing integers from 1 to n - 1, inclusive, with exactly one repeated. Describe a fast algorithm for finding the integer in A that is repeated.

O(n) for time and no extra space:

public class findDuplicate {
	
	public static int findDup(int A[]){
		
		int dupNum = 0;
		for(int i = 0; i < A.length; i++){
			dupNum += A[i] - i;
		}
		return dupNum;
		
	}
	
	public static void main(String args[]){
		
		int[] A = {5,1,2,3,2,4};
		System.out.println(findDup(A));
	}

}

O(n) for time and extra space:

public class findDuplicate {
	
	public static int findDup(int A[]){
		
		int B[] = new int[A.length];	
		for(int i = 0; i < B.length; i++){
			B[i] = 0;
		}
		for(int j = 0; j < A.length; j++){
			if(B[A[j]] == 0){
				B[A[j]] = 1;
			}else{
				return A[j];
			}	
		}
		return 0;
		
	}
	
	public static void main(String args[]){
		
		int[] A = {5,1,2,3,2,4};
		System.out.println(findDup(A));
	}

}