LeetCode 564. Find the Closest Palindrome
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原题链接在这里:https://leetcode.com/problems/find-the-closest-palindrome/
题目:
Given an integer n, find the closest integer (not including itself), which is a palindrome.
The ‘closest‘ is defined as absolute difference minimized between two integers.
Example 1:
Input: "123" Output: "121"
Note:
- The input n is a positive integer represented by string, whose length will not exceed 18.
- If there is a tie, return the smaller one as answer.
题解:
First find all the candidates, add "9999..999", "10000..0001".
Add first half, + reverse of (first half).
e.g. 12567, its first half is 125, the candiates couldbe 12521, 12421, 12621.
Then check all the candidates and find the closest one.
Time Complexity: O(mn). m = is candidates list size. n is candiate length.
Space: O(mn).
AC Java:
1 class Solution { 2 public String nearestPalindromic(String n) { 3 if(n == null || n.length() == 0){ 4 return n; 5 } 6 7 int len = n.length(); 8 if(len == 1){ 9 int val = Integer.valueOf(n); 10 return val > 0 ? "" + (val - 1) : "" + (val + 1); 11 } 12 13 List<String> cans = new ArrayList<>(); 14 15 cans.add(allNine(len - 1)); 16 cans.add(oneZero(len + 1)); 17 18 int halfLen = (len + 1) / 2; 19 String sub = n.substring(0, halfLen); 20 if(len % 2 == 1){ 21 cans.add(sub + new StringBuilder(sub).deleteCharAt(halfLen - 1).reverse().toString()); 22 long halfVal = Long.valueOf(sub); 23 long plusOne = halfVal + 1; 24 cans.add("" + plusOne + new StringBuilder("" + plusOne / 10).reverse().toString()); 25 long minusOne = halfVal - 1; 26 if(minusOne > 0){ 27 cans.add( "" + minusOne + new StringBuilder("" + minusOne / 10).reverse().toString()); 28 } 29 }else{ 30 cans.add(sub + new StringBuilder(sub).reverse().toString()); 31 long halfVal = Long.valueOf(sub); 32 long plusOne = halfVal + 1; 33 cans.add("" + plusOne + new StringBuilder("" + plusOne).reverse().toString()); 34 long minusOne = halfVal - 1; 35 if(minusOne > 0){ 36 cans.add("" + minusOne + new StringBuilder("" + minusOne).reverse().toString()); 37 } 38 } 39 40 long diff = Long.MAX_VALUE; 41 String res = ""; 42 long nValue = Long.valueOf(n); 43 for(String can : cans){ 44 if(can.equals(n)){ 45 continue; 46 } 47 48 long canValue = Long.valueOf(can); 49 if(Math.abs(canValue - nValue) < diff){ 50 diff = Math.abs(canValue - nValue); 51 res = can; 52 }else if(Math.abs(canValue - nValue) == diff && (res.length() == 0 || canValue < Long.valueOf(res))){ 53 res = can; 54 } 55 } 56 57 return res; 58 } 59 60 private String allNine(int len){ 61 char [] charArr = new char[len]; 62 Arrays.fill(charArr, ‘9‘); 63 return new String(charArr); 64 } 65 66 private String oneZero(int len){ 67 char [] charArr = new char[len]; 68 Arrays.fill(charArr, ‘0‘); 69 charArr[0] = ‘1‘; 70 charArr[len - 1] = ‘1‘; 71 return new String(charArr); 72 } 73 }
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