# 【LeetCode】Reverse Integer(整数反转)

### 题目描述：

输入: 123



示例 2:

输入: -123



输入: 120



• INT_MAX_VALUE = 2^31-1 = 2147483647
• INT_MIN_VALUE = -2^31 = -2147483648

### 提交代码：

class Solution {
public int reverse(int x) {
int res=0;
//boolean flag=x<0?false:true;
if(x>0){
while(true){
res+=x%10;
x/=10;
if(x==0)return res;
if(res>214748364||res==214748364&&x%10>7)return 0;
res*=10;
}
}else{
while(true){
res+=x%10;
x/=10;
if(x==0)return res;
if(res<-214748364||res==-214748364&&x%10<-8)return 0;
res*=10;
}
}
}
}

class Solution {
public int reverse(int x) {
int res=0;
//boolean flag=x<0?false:true;
while(true){
res+=x%10;
x/=10;
if(x==0)return res;
if(res>214748364||res==214748364&&x%10>7)return 0;
if(res<-214748364||res==-214748364&&x%10<-8)return 0;
res*=10;
}
}
}

### 个人总结：

StringBuffer 法：

class Solution {
public int reverse(int x) {
StringBuffer sb = new StringBuffer();
String str = sb.append(String.valueOf(x < 0 ? x * -1 : x)).reverse().toString();
try{
return x < 0 ? Integer.parseInt(str) * -1 : Integer.parseInt(str);
}catch(Exception e){
return 0;
}
}
}

class Solution {
public int reverse(int x) {
String s=String.valueOf(Math.abs(x));
StringBuilder sb=new StringBuilder();
for (int i=s.length()-1;i>=0;--i){
sb.append(s.charAt(i));
}
try{
return x>0?Integer.valueOf(sb.toString()):-Integer.valueOf(sb.toString());
}catch (Exception e){
return 0;
}
}
}

class Solution {
public int reverse(int x) {
char[] s = Integer.toString(x).toCharArray();
int f = 0,b = s.length-1;
String string = "";
if (s[f] == '-') {
string += '-';
++f;
}
while (b >= 0&& s[b] == '0') {
--b;
}
while (f <= b) {
string += s[b--];
}
if (string != "") {
long num = Long.valueOf(string);
if (num < (1<<31)-1 && num > -(1<<31)) {
return (int)num;
}
}
return 0;
}
}

posted @ 2019-04-05 15:11  1000sakura  阅读(75)  评论(0编辑  收藏