1、字符串倒叙排列。
Write a function that takes a string as input and returns the string reversed.
Example 1:
Input: "hello"
Output: "olleh"
Example 2:
Input: "A man, a plan, a canal: Panama"
Output: "amanaP :lanac a ,nalp a ,nam A"//方法一
class Solution {
public String reverseString(String s) {
char temp;
int n = s.length();
char[] myChar=s.toCharArray();//不用指定长度？字符串指定长度？
for(int i=0;i<n/2;i++)//小心错误，不能为 i<n，则交换了两次又交换成了hello
{
temp=myChar[i];
myChar[i]=myChar[n-i-1];
myChar[n-i-1]=temp;
}
return new String(myChar);
}
}2、整数倒叙问题。
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.//方法
class Solution {
public int reverse(int x) {
int result = 0;
while(x!=0)
{
int tail = x%10;
int newResult = result*10+tail;
if((newResult-tail)/10!=result)//tail 可以省略 因为newResult/10已经去掉了末位
{
return 0;
}
result = newResult;
x=x/10;
}
return result;
}
}
as many others, I didn't get the line if ((newResult - tail) / 10 != result).
We're interested on what happens when an integer overflows. Well, it is rolled over. Practically speaking, if you would try
public static void main(String[] args) {
int rollMeOver= Integer.MAX_VALUE + 1;
System.out.println(rollMeOver);
}
You will get as an output -2147483648 which represents the lowest value for an integer (Integer.MIN_VALUE).
Thus, in our problem, if newResult is going to overflow we are sure that newResult / 10 != result (this is the reason that @Inception_wzd said that we don't need to subtract the tail first because by / 10 we are already losing the last digit).
By the way, the same thing happens for the underflow.
public static void main(String[] args) {
int rollMeOver= Integer.MIN_VALUE - 1;
System.out.println(rollMeOver);
}
This is going to output the Integer.MAX_VALUE which is 2147483647 .3、寻找字符串中第一个不重复的字符。
Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
Examples:
s = "leetcode"
return 0.
s = "loveleetcode",
return 2.
Note: You may assume the string contain only lowercase letters.//常规解法 o（n^2）
class Solution {
public int firstUniqChar(String s) {
char [] myChar = s.toCharArray();
int n = s.length();
int j=0;
while(j<n)
{
for(int i=0;i<n;i++)
{
if ((myChar[j]!=myChar[i])&&(i!=j))
return j;
}
j++;
}
return -1;
}
}
//解法二：
//Get the frequency of each character.
//Get the first character that has a frequency of one.
//Actually the code below passes all the cases. However, according to @xietao0221, we //could change the size of the frequency array to 256 to store other kinds of characters.
public class Solution {
public int firstUniqChar(String s) {
int freq [] = new int[26];
for(int i = 0; i < s.length(); i ++)
freq [s.charAt(i) - 'a'] ++;
for(int i = 0; i < s.length(); i ++)
if(freq [s.charAt(i) - 'a'] == 1)
return i;
return -1;
}
}