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Copy pathReverse_Integer_7.java
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60 lines (53 loc) · 1.64 KB
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package leet.code;
/**
* Example1: x = 123, return 321
Example2: x = -123, return -321
*
* If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
* Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows.
* How should you handle such cases?
*/
public class Reverse_Integer_7 {
public static void main(String[] args) {
//System.out.println(reverse1(542000)); // 这种做法不好 没有考虑 000
System.out.println(reverse2(1012034));
}
/**
*
* method 1
*/
public static int reverse1(int x) {
if (Integer.valueOf(x).toString().toCharArray().length > 9 ) {
return 0;
}
char[] temp = removeTail0(Integer.valueOf(x).toString()).toCharArray();
String result = "";
for (int a = temp.length -1; a >= 0; a--) { // 反着遍历
result += temp[a];
}
return Integer.valueOf(result);
}
// 去除末尾 0
public static String removeTail0(String str) {
if(!str.substring(str.length() -1).equals("0")) {
return str;
}
return removeTail0(str.substring(0, str.length() -1));
}
/**
* method 2 ，高手算法
*/
public static int reverse2(int x) {
int result = 0;
while (x != 0) {
int tail = x % 10; // 算法不停地求个位 = 求余数
int newResult = result * 10 + tail; // 倒序一次移一位4 34 234 ...
if ((newResult - tail) / 10 != result) {
return 0;
}
result = newResult;
x = x / 10; // 秒 /10 就等于 倒着减一位， 厉害
}
return result;
}
}