验证回文字符串
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Note: For the purpose of this problem, we define empty string as valid palindrome.
Example 1:
Input: “A man, a plan, a canal: Panama” Output: true
Example 2:
Input: “race a car” Output: false
easy one. 字符串练手。
class Solution {
public boolean isPalindrome(String s) {
if(s.length() == 0)
return true;
ArrayList<Character> letter = new ArrayList<>();
for(char ch:s.toLowerCase().toCharArray()){
if((ch >= 'a'&&ch <= 'z')||(ch >= 'A'&&ch <= 'Z')||(ch >= '0'&&ch <='9'))
letter.add(ch);
}
int len = letter.size();
for(int i = 0; i < len / 2; i++){
if(letter.get(i) != letter.get(len - i -1))
return false;
}
return true;
}
}
数数并说
报数序列是指一个整数序列，按照其中的整数的顺序进行报数，得到下一个数。其前五项如下：
1 1
2 11
3 21
4 1211
5 111221
1 被读作 “one 1” (“一个一”) , 即 11。 11 被读作 “two 1s” (“两个一”）, 即 21。 21 被读作 “one 2”, “one 1” （”一个二” , “一个一”) , 即 1211。
给定一个正整数 n ，输出报数序列的第 n 项。
注意：整数顺序将表示为一个字符串。
示例 1:
输入: 1 输出: “1”
示例 2:
输入: 4 输出: “1211”
首先需要理解题目。对字符串中连续相同的子字符串合并成「数量+值」，单个的则写成「”1”+值」。然后迭代进行，直到循环n-1次达到目标。
class Solution {
int pointer = 0;
String tempResult = "";
int amount = 0;
public String countAndSay(int n) {
String result = "1";
if(n==1)
return result;
while(n>1) {
while(pointer < result.length()-1) {//count next sequence
analyze(result);//read out loud!
}
if(pointer==result.length()-1) {//case like 21 at end
tempResult += "1";
tempResult += result.charAt(result.length()-1);
}
//case like 21 || 11 at end
result = tempResult;
n--;
pointer = 0;
tempResult ="";
}
return result;
}
public void analyze(String result) {
for(amount = 1;pointer <result.length()-1&&
result.charAt(pointer)==result.charAt(pointer+1); ) {
amount++;
pointer++;
}
tempResult += Integer.toString(amount);
tempResult += result.charAt(pointer);
pointer++;
}
}
最长公共前缀
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string “”.
Example 1:
Input: [“flower”,”flow”,”flight”] Output: “fl”
Example 2:
Input: [“dog”,”racecar”,”car”] Output: “” Explanation: There is no common prefix among the input strings.
Note:
All given inputs are in lowercase letters a-z.
要先计算最短字符串有多长，处理一些特例，然后再查找相同字符（感觉储存字符串数组，有点像储存二维数组啊）。
class Solution {
public String longestCommonPrefix(String[] strs) {
int size = strs.length;
if(size==0)
return "";
if(size==1)
return strs[0];
else{
int min = strs[0].length();
for(int i = 0; i < size-1; i++){
if(min > strs[i+1].length())
min = strs[i+1].length();//know how short could a str be
}
if(min==0)
return "";
for(int k = 0; k < min; k++){
for(int j = 0; j < size-1; j++){
if(strs[j].charAt(k)!=strs[j+1].charAt(k)){
if(k==0)
return "";//no common prefix
j = size;//this is just a flag
return strs[0].substring(0,k);
}
}
}
return strs[0].substring(0,min);
}
}
}