125 Valid Palindrome Easy

Given a string, determine if it is a palindrome（回文）, considering only alphanumeric（字母+数字） characters and ignoring cases（忽略大小写）.

For example,

"A man, a plan, a canal: Panama" is a palindrome.

"race a car" is not a palindrome.

Note:

Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

思路：从两头向中间依次对应比较。

class Solution {
    public boolean isPalindrome(String s) {
        if(s == "") {
            return true;
        }
        //所有字母变为小写
        s = s.toLowerCase();    
        //删除字母数字外的其它字符
        s = s.replaceAll("[^a-z0-9]","");   
        int n = s.length();
        int i = 0, j = n-1;
        while(i<j) {
            if(s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;            
        }
        return true;
    }
}

680 Valid Palindrome II Easy

Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.

Example 1:

Input: "aba"

Output: True

Example 2:

Input: "abca"

Output: True

Explanation: You could delete the character 'c'.

Note:

The string will only contain lowercase characters a-z. The maximum length of the string is 50000.

思路：从两头开始比较，当遇到两个字符不相同时，会有两种情况：

将左边一个舍去，剩下的都对应相同；

将右边一个舍去，剩下的都对应相同。

这两种情况要分别尝试！不能在一个循环里同时试。

舍去的时候，看一下舍去后是否相同，如果是，继续，如果不是就说明舍去这个行不通。

为什么说舍左边/舍右边不能在一个循环里同时试呢？因为你在同一个循环里试的时候肯定有个先后，而存在一种情况，在当下，舍左&舍右下一步都是相同的，这时候你肯定就舍左了，因为舍左判断在前，结果到后面又发现行不通。 但其实舍右是行得通的，而程序里就忽略了这种情况！

例子："mlcupuufxxfuupuculm"

class Solution {
    public boolean validPalindrome(String s) {
        int i = 0, j = s.length()-1;
        boolean hasDeleted = false;
        boolean leftFail = false;
        while(i<j) {
            if(s.charAt(i)!=s.charAt(j)) {
                if(!hasDeleted) {
                    if(j-i == 1) {
                        return true;
                    }
                    if(s.charAt(i+1)==s.charAt(j)) {
                        i++;
                        hasDeleted = true;
                    } else {
                        leftFail = true;
                        break;  
                    }
                } else {
                    leftFail = true;
                    break;                    
                }                
            }
            i++;
            j--;
        }
        if(!leftFail) {
            return true;
        } else {
            hasDeleted = false;
            i = 0;
            j = s.length()-1;
            while(i<j) {
                if(s.charAt(i)!=s.charAt(j)) {
                    if(!hasDeleted) {
                        if(j-i == 1) {
                            return true;
                        }
                        if(s.charAt(i)==s.charAt(j-1)) {
                            j--;
                            hasDeleted = true;
                        } else {
                            return false;
                        }
                    } else {
                        return false;
                    }                
                }
                i++;
                j--;
            }
            return true;
        }
    }
}

67 Add Binary Easy

Given two binary strings, return their sum (also a binary string).

For example,

a = "11"

b = "1"

Return "100".

class Solution {
    public String addBinary(String a, String b) {
        int carry = 0;
        int i = a.length() - 1;
        int j = b.length() - 1;
        String result = "";
        while(i>=0 && j>=0) {
            // char -> int
            int A = a.charAt(i)-'0';
            int B = b.charAt(j)-'0';
            int temp = A+B+carry;
            int res = temp%2;       // k进制就用k
            carry = temp/2;
            result = res + result;            
            i--;
            j--;
        }
        if(i>=0) {
            while(i>=0) {
                int A = a.charAt(i)-'0';            
                int temp = A+carry;
                int res = temp%2;
                carry = temp/2;
                result = res + result;            
                i--;
            }            
        } else if(j>=0) {
            while(j>=0) {
                int B = b.charAt(j)-'0';            
                int temp = B+carry;
                int res = temp%2;
                carry = temp/2;
                result = res + result;            
                j--;
            } 
        }
        if(carry == 1) {
            return "1"+result;      //字符串拼接
        } else {
            return result;
        }
    }
}

49 Group Anagrams Medium

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],

Output:

[

["ate","eat","tea"],

["nat","tan"],

["bat"]

]

Note:

All inputs will be in lowercase.

The order of your output does not matter.

思路：用散列表，Key是按字母序升序的字符串，Value是String的ArrayList，放每一个原始字符串。遍历字符串数组，对每一个字符串，取得其排序后的字符串，然后将其插入散列表。

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String,ArrayList<String>> map = new HashMap<>();
        for(int i=0; i<strs.length; i++) {
            String sortedStr = sortStr(strs[i]);
            if(map.containsKey(sortedStr)) {
                map.get(sortedStr).add(strs[i]);
            } else {
                ArrayList<String> list = new ArrayList();
                list.add(strs[i]);
                map.put(sortedStr,list);
            }
        }
        List<List<String>> result = new ArrayList<>();
        map.forEach((key,value) -> {
            result.add(value);
        });
        return result;
    }
    private String sortStr(String str) {
        char[] charArray = str.toCharArray();
        Arrays.sort(charArray);
        return new String(charArray);        
    }
}