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7. Reverse Integer
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.
解题分析
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2018-07-01
*/
public static int reverse(int x) {
int sign = 1;
if (x < 0) {
sign = -1;
if (x < -Integer.MAX_VALUE) {
return 0;
}
x = -x;
}
int result = 0;
while (x > 0) {
int digit = x % 10;
int num = result * 10 + digit;
if ((num - digit) / 10 != result) {
return 0;
}
result = num;
x /= 10;
}
return sign * result;
}
public static int reverse1(int x) {
if (x == 0
|| x > Math.pow(2, 31)
|| x < -Math.pow(2, 31)) {
return 0;
}
int sign = 1;
int positiveNum = x;
if (x < 0) {
sign = -1;
positiveNum = x * sign;
}
boolean zeroOfEnd = true;
List<Integer> bitNums = new ArrayList<>(25);
for (int i = positiveNum; i > 0; ) {
int bitNum = i % 10;
i = i / 10;
if (zeroOfEnd && bitNum == 0) {
continue;
}
bitNums.add(bitNum);
if (zeroOfEnd) {
zeroOfEnd = false;
}
}
long result = 0;
for (int j = 0; j < bitNums.size(); j++) {
result += bitNums.get(j) * ((long) Math.pow(10, bitNums.size() - j - 1));
}
if (result > Integer.MAX_VALUE) {
return 0;
}
return (int) result * sign;
}