給一個 32bit 有正負號的整數，返回反轉過後的數字。
LeetCode 題目連結
https://leetcode.com/problems/reverse-integer
題目
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123 Output: 321
Example 2:
Input: -123 Output: -321
Example 3:
Input: 120 Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Accept 作法
要考慮溢位問題，還有負號永遠擺前面，以及遇到 0 在前面反轉後都不顯示。
Runtime: 2 ms
Memory: 36.9 MB
Java
Runtime: 2 ms
Memory: 36.9 MB
class Solution {
class Solution {
public int reverse(int x) {
String intStr = String.valueOf(x);
StringBuilder result = new StringBuilder();
boolean isCons = false;
for(int i = 0; i< intStr.length();i++){
if(intStr.charAt(i) == '-'){
result.append('-');
isCons = true;
}else{
if(isCons){
if(intStr.charAt(intStr.length() - i) != '-'){
result.append(intStr.charAt(intStr.length() - i));
}
}else{
result.append(intStr.charAt(intStr.length() - 1 - i));
}
}
}
try{
int resultInt = Integer.valueOf(result.toString());
return resultInt;
}catch(Exception e){
return 0;
}
}
}
}
官方的答案更簡單。我果然還是想的太冗長了。
class Solution {
public int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
rev = rev * 10 + pop;
}
return rev;
}
}
更多 LeetCode 相關資源
複習程式面試書籍
除了 LeetCode 練習外，我也入手了這本，題庫來自真正的面試，並非摘自教科書。它們反映出頂尖公司真正會出的題目，你可以藉此做好充分準備。
需要的話可以看看，寫得很仔細。
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