Đưa ra một số, tìm palindrom nhỏ nhất tiếp theo lớn hơn số này. Ví dụ: nếu số đầu vào là là 2 3 5 4 5, thì đầu ra phải là 2 3 3 6 3 2. Và nếu số đầu vào là trên 9 9 9 9, thì đầu ra phải là 1 0 0 0 1. & Nbsp; đầu vào được coi là một mảng. Mỗi mục trong mảng đại diện cho một chữ số trong số đầu vào. Đặt mảng là ‘num [] và kích thước của mảng là’ n,
The input is assumed to be an array. Every entry in array represents a digit in input number. Let the array be ‘num[]’ and size of array be ‘n’
Có thể có ba loại đầu vào khác nhau cần được xử lý riêng. & NBSP; 1] Số đầu vào là palindrom và có tất cả 9s. Ví dụ, 9 9 9 9 9. Đầu ra phải là 1 0 0 0 1 và NBSP; 2] Số đầu vào không phải là palindrom. Ví dụ: 1 2 3 3 4. Đầu ra phải là 1 3 3 1 1 & NBSP; 3] Số đầu vào là palindrom và không có tất cả 9s. Ví dụ: 1 2 2 1 1. Đầu ra phải là 1 3 3 3 1.
1]
The input number is palindrome and has all 9s. For example “9 9 9”. Output should be “1 0 0 1”
2] The input number is not palindrome. For example “1 2 3 4”. Output should be “1 3 3 1”
3] The input number is palindrome and doesn’t have all 9s. For example “1 2 2 1”. Output should be “1 3 3 1”.
Giải pháp cho đầu vào loại 1 là dễ dàng. Đầu ra chứa N + 1 chữ số trong đó các chữ số góc là 1 và tất cả các chữ số giữa các chữ số góc là 0. Bây giờ, trước tiên chúng ta hãy nói về đầu vào loại 2 và 3. Làm thế nào để chuyển đổi một số đã cho thành một palindrom lớn hơn? Để hiểu giải pháp, trước tiên chúng ta hãy xác định hai thuật ngữ sau: & nbsp;
Now let us first talk about input type 2 and 3. How to convert a given number to a greater palindrome? To understand the solution, let us first define the following two terms:
Bên trái: Nửa bên trái của số đã cho. Bên trái của người Viking 1 2 3 4 5 6, là 1 2 3 3 và bên trái của 1 2 3 4 5 5 là 1 2
Bên phải: Nửa bên phải của số đã cho. Bên phải của người Viking 1 2 3 4 5 6, là 4 5 6 và bên phải của 1 2 3 4 5
Để chuyển đổi thành palindrom, chúng ta có thể lấy gương của bên trái của nó hoặc lấy gương của bên phải của nó. Tuy nhiên, nếu chúng ta lấy gương của phía bên phải, thì palindrom nên hình thành không được đảm bảo là palindrom lớn hơn tiếp theo. Vì vậy, chúng ta phải lấy gương của bên trái và sao chép nó sang bên phải. Nhưng có một số trường hợp phải được xử lý theo những cách khác nhau. Xem các bước sau đây. Chúng tôi sẽ bắt đầu với hai chỉ số i và j. Tôi chỉ vào hai yếu tố giữa [hoặc chỉ vào hai yếu tố xung quanh phần tử giữa trong trường hợp N là lẻ]. Chúng tôi từng người di chuyển tôi và J xa nhau.
We will start with two indices i and j. i pointing to the two middle elements [or pointing to two elements around the middle element
in case of n being odd]. We one by one move i and j away from each other.
Bước 1. Ban đầu, bỏ qua phần của phía bên trái giống như phần tương ứng của phía bên phải. Ví dụ: nếu số đó là 8 3 4 2 2 4 6 9, chúng ta sẽ bỏ qua bốn phần tử giữa. Bây giờ tôi chỉ vào phần tử 3 và j bây giờ chỉ vào phần tử 6. Initially, ignore the part of left side which is same as the corresponding part of right side. For example, if the number is “8 3 4 2 2 4 6 9″, we ignore the middle four elements. i now points to element 3 and j now points to element 6.
Bước 2. Sau bước 1, các trường hợp sau phát sinh: Trường hợp 1: Chỉ số I & J vượt qua ranh giới. & NBSP; Trường hợp này xảy ra khi số đầu vào là palindrom. Trong trường hợp này, chúng tôi chỉ cần thêm 1 vào chữ số giữa [hoặc các chữ số trong trường hợp N là chẵn] tuyên truyền mang theo chữ số MSB của bên trái và sao chép gương của bên trái sang bên phải. & NBSP; Số được cho là 1 2 9 2 1, chúng tôi tăng 9 đến 10 và tuyên truyền mang theo. Vì vậy, con số trở thành "1 3 0 3 1" After step 1, following cases arise:
Case 1: Indices i & j cross the boundary.
This
case occurs when the input number is palindrome. In this case, we just add 1 to the middle digit [or digits in case n is even] propagate the carry towards MSB digit of left side and simultaneously copy mirror of the left side to the right side.
For example, if the given number is “1 2 9 2 1”, we increment 9 to 10 and propagate the carry. So the number becomes “1 3 0 3 1”
Trường hợp 2: Có các chữ số bên trái giữa bên trái và bên phải không giống nhau. Vì vậy, chúng tôi chỉ phản chiếu phía bên trái sang bên phải và cố gắng giảm thiểu số được hình thành để đảm bảo palindrom nhỏ nhất tiếp theo. & Nbsp; trong trường hợp này, có thể có hai trường hợp phụ. & Nbsp; There are digits left between left side and right side which are not same. So, we just
mirror the left side to the right side & try to minimize the number formed to guarantee the next smallest palindrome.
In this case, there can be two sub-cases.
2.1] Sao chép phía bên trái sang bên phải là đủ, chúng tôi không cần phải tăng bất kỳ chữ số nào và kết quả chỉ là gương của bên trái. Sau đây là một số ví dụ về trường hợp phụ này. 5 2 1 và & nbsp; palindrom tiếp theo cho nhóm 1 4 5 8 7 6 7 8 3 2 2 ″ là 1 4 5 8 7 6 7 8 5 4 1 1 & NBSP; Làm thế nào để chúng ta kiểm tra trường hợp phụ này? Tất cả những gì chúng ta cần kiểm tra là chữ số ngay sau phần bị bỏ qua trong Bước 1. Chữ số này được tô sáng trong các ví dụ trên. Nếu chữ số này lớn hơn chữ số tương ứng theo chữ số bên phải, thì việc sao chép bên trái sang bên phải là đủ và chúng tôi không cần phải làm bất cứ điều gì khác. Copying the left side to the right side is sufficient, we don’t need to increment any digits and the result is just mirror of left side. Following are some examples of this sub-case.
Next palindrome for “7 8 3 3 2 2″ is “7 8 3 3 8 7”
Next
palindrome for “1 2 5 3 2 2″ is “1 2 5 5 2 1”
Next palindrome for “1 4 5 8 7 6 7 8 3 2 2″ is “1 4 5 8 7 6 7 8 5 4 1”
How do we check for this sub-case? All we need to check is the digit just after the ignored part in step 1. This digit is highlighted in above examples. If this digit is greater than the corresponding digit in right side digit, then copying the left side to the right side is sufficient and we don’t need to do anything else.
2.2] Sao chép bên trái sang bên phải là không đủ. Điều này xảy ra khi chữ số được xác định ở trên của bên trái nhỏ hơn. Sau đây là một số ví dụ về trường hợp này. 3 2 1 1 & nbsp; palindrom tiếp theo cho nhóm 9 4 1 8 7 9 7 8 3 2 2 ″ là 9 4 1 8 8 8 8 1 4 9 đến chữ số giữa [hoặc các chữ số trong trường hợp N chẵn] tuyên truyền số lượng mang về phía MSB của bên trái và đồng thời sao chép gương của phía bên trái sang bên phải.Copying the left side to the right side is NOT sufficient. This happens when the above defined digit of left side is smaller. Following are some examples of this case.
Next palindrome for “7 1 3 3 2 2″ is “7 1 4 4 1 7”
Next palindrome for “1 2 3 4 6 2 8″ is “1 2 3 5 3 2 1”
Next palindrome for “9 4 1 8 7 9 7 8 3 2 2″ is “9 4 1 8 8 0 8 8 1 4 9”
We handle this subcase like Case 1. We just
add 1 to the middle digit [or digits in case n is even] propagate the carry towards MSB digit of left side and simultaneously copy mirror of the left side to the right side.
Cách tiếp cận 1: Cách tiếp cận cơ bản để tìm số palindrom nhỏ nhất tiếp theo.Basic Approach for Finding the next smallest Palindrome Number.
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56Next palindrome is:9 4 1 8 8 0 8 8 1 4 986
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56Next palindrome is:9 4 1 8 8 0 8 8 1 4 986
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36Next palindrome is:9 4 1 8 8 0 8 8 1 4 91
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33Next palindrome is:9 4 1 8 8 0 8 8 1 4 9686
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06Next palindrome is:9 4 1 8 8 0 8 8 1 4 91
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38Next palindrome is:9 4 1 8 8 0 8 8 1 4 91
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