The value range is 0 to 1001 ~ 1000 of the numbers to find all the repeated numbers. No additional space is needed on the basis of O (n ).

I saw a question. "The 1001 ranges from 1 ~ 1000 of the number, find a repeated number, O (n) based on no additional space "someone gives the solution: int I = 0; Int T = 0; While (A [I]! = 0 ){ T = A [I]; A [I] = 0; I = T; } System. Out. println (t); in addition, to meet the "no extra space required" requirement, remove t to this: int I = 0; while (A [I]! = 0) I = A [I] + (A [I] = 0); system. Out. println (I); yes, this solution can solve the problem. However, to increase the difficulty, if I change the number range to 0 ~ 1000, this AlgorithmNo, even if you change the algorithm to int I = 0; while (A [I]! =-1) I = A [I] + 1 + (A [I] =-1); system. out. println (I); // output I; When a [0] = 0, the problem cannot be solved. If you change the algorithm to int I = 0; while (A [I]! =-1) I = A [I] + 1 + (A [I] =-1) + 1; system. out. println (I); // output I; the array subscript is out of bounds. To increase the difficulty, the new question is: the range of 1001 is 0 ~ 1000To find out AllThere is no extra space for repeated numbers based on O (N). Can you come up with a solution? I already have the solution. I will not announce it for the moment.