力扣 java题
# 简单 [力扣](https://leetcode-cn.com/problems/reverse-integer/)
## 两数之和
```
public static int[] wtl(){
int[] nums ={2, 7, 11, 15};
int target = 9;
Map<Integer,Integer> map=new HashMap<Integer,Integer>();
for (int i=0 ;i<nums.length;++i ){
if(map.containsKey(target-nums[i])){
return new int[]{map.get(target-nums[i]),i};
}
map.put(nums[i],i);
}
return null;
}
```
> map.containsKey 比较map里的key是否存在；
## 整数反转
```
public static int reverse(int x) {
int wtl=0;
while (x!=0){
if(wtl*10/10!=wtl){
wtl=0;
break;
}
wtl=wtl*10+x%10;
x=x/10;
}
return wtl;
}
```
## 回文数
```
public static boolean reverse(int x) {
if (x < 0) {
return false;
}
int w = 0;
int t=x;
while (x != 0) {
w= w * 10 + x % 10;
x=x/10;
}
return w == t;
}
```
## 罗马数字转整数
```
public static int reverse2(String s){
int sum =0; //总值
int presentS=getValue(s.charAt(0)); //当前值
int presentNextS=0; //下一个值
for (int i = 1; i <s.length() ; i++) {
presentNextS=getValue(s.charAt(i));//获取下一个值
if (presentS<presentNextS){ //当前值<下一个值
sum-=presentS;
}else {
sum+=presentS;
}
presentS=presentNextS; //赋值（当前值=下一个值）
}
sum+=presentS; //加上最后一个值；
return sum;
}
public static int getValue(char ch) {
switch(ch) {
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return 0;
}
}
```
## 最长公共前缀
```
public static String reverse3(String[] strs){ //纵向查找
if (strs == null || strs.length == 0) {
return "";
}
char c=0;
for (int i = 0; i < strs[0].length() ; i++) {
c=strs[0].charAt(i);
for (int j = 1; j < strs.length ; j++) {
if(strs[j].length()==i || c!=strs[j].charAt(i)){
return strs[0].substring(0,i);
}
}
}
return strs[0];
}
```