String

4月28号

1  5 Longest Palindromic Substring   分奇偶，找最大

    public String longestPalindrome(String s) {
        int n = s.length();
        if (n < 2) return s;
        String res = "";
        for (int i = 0; i < n * 2 - 1; i++)
        {
            int l = i / 2, r = i / 2;
            if (i % 2 == 1)
            {
                r++;
            }
            String cur = longest(s,l,r);
            if (res.length() < cur.length())
            {
                res = cur;
            }
        }
        return res;
    }
    public String longest(String s, int l, int r)
    {
        while (l >=0 && r < s.length() && s.charAt(l) == s.charAt(r))
        {
            l--;r++;
        }
        return s.substring(l+1,r);
    }

2 6 ZigZag Conversion     n个sb 下上遍历

    public String convert(String s, int numRows) {
        char[] c = s.toCharArray();
        int len = c.length;
        StringBuffer[] sb = new StringBuffer[numRows];
        for (int i = 0; i < numRows; i++)
        {
            sb[i] = new StringBuffer();
        }
        int i = 0;
        while (i < len)
        {
            for (int ind = 0; ind < numRows && i < len; ind++)
            {
                sb[ind].append(c[i++]);
            }
            for (int ind = numRows - 2; ind >= 1 && i < len; ind--)
            {
                sb[ind].append(c[i++]);
            }
        }
        for (int in = 1; in < numRows; in++)
        {
            sb[0].append(sb[in]);
        }
        return sb[0].toString();
    }

3  10 Regular Expression Match

public boolean isMatch(String s, String p) {

    if (s == null || p == null) {
        return false;
    }
    boolean[][] dp = new boolean[s.length()+1][p.length()+1];
    dp[0][0] = true;
    for (int i = 0; i < p.length(); i++)
    {
        if (p.charAt(i) == '*' && dp[0][i - 1])
        {
            dp[0][i + 1] = true;
        }
    }
    for (int i = 0; i < s.length(); i++)
    {
        for (int j = 0; j < p.length(); j++)
        {
            if (p.charAt(j) == '.')
            {
                dp[i+1][j + 1] = dp[i][j];
            }
            if (p.charAt(j) == s.charAt(i))
            {
                dp[i+1][j + 1] = dp[i][j];
            }
            if (p.charAt(j) == '*')
            {
                if (p.charAt(j - 1) != s.charAt(i) && p.charAt(j - 1) != '.')
                {
                    dp[i+1][j+1]=dp[i+1][j-1];
                }
                else
                {
                    dp[i+1][j+1]=(dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
                }
            }
        }
    }
     return dp[s.length()][p.length()];
}

4 14 Longest Common prefix

    public String longestCommonPrefix(String[] strs) {
    if(strs == null || strs.length == 0)    return "";
    String pre = strs[0];
    int i = 1;
    while(i < strs.length){
        while(strs[i].indexOf(pre) != 0)
            pre = pre.substring(0,pre.length()-1);
        i++;
    }
    return pre;
    }

 4 月 29 号

5  17 Letter Combination of a Phone number   维护队列   ，遍历数字 

    public List<String> letterCombinations(String digits) {
        LinkedList<String> queue = new LinkedList<String>();
        if (digits == null || digits.length() == 0) return queue;
        queue.add("");
        String[] map = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        for (int i = 0; i < digits.length(); i++)
        {
            int x = Character.getNumericValue(digits.charAt(i));
            while (queue.peek().length() == i)
            {
                String t = queue.remove();
                for(char c : map[x].toCharArray())
                {
                    queue.add(t + c);
                }
            }
        }
    return queue;
    }

6 20 Valid parentheses   维护堆  

    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<>();
        for (char c : s.toCharArray())
        {
            if (c == '(')
            {
                stack.push(')');
            }
            else if (c == '{')
            {
                stack.push('}');
            }
            else if (c == '[')
            {
                stack.push(']');
            }
            else if (stack.isEmpty() || c != stack.pop())
            {
                return false;
            }
        }
        return stack.isEmpty();
    }

7 22 Generate Parentheses  维护栈 ，开括号加  毕小于开

    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<>();
        help(n, "", 0, 0, res);
        return res;
    }
    public void help(int n, String item, int open, int close, List<String> res)
    {
        if (item.length() == n * 2)
        {
            res.add(item);
        }
        if (open < n)
        {
            help(n, item + "(", open + 1, close, res);
        }
        if (close < open)
        {
            help(n, item + ")", open, close + 1, res);
        }
    }

8 30 Substring with Concatenation of all words   hashmap   遍历

public static List<Integer> findSubstring(String S, String[] L) {
    List<Integer> res = new ArrayList<Integer>();
    if (S == null || L == null || L.length == 0) return res;
    int len = L[0].length(); // length of each word
    
    Map<String, Integer> map = new HashMap<String, Integer>(); // map for L
    for (String w : L) map.put(w, map.containsKey(w) ? map.get(w) + 1 : 1);
    
    for (int i = 0; i <= S.length() - len * L.length; i++) {
        Map<String, Integer> copy = new HashMap<String, Integer>(map);
        for (int j = 0; j < L.length; j++) { // checkc if match
            String str = S.substring(i + j*len, i + j*len + len); // next word
            if (copy.containsKey(str)) { // is in remaining words
                int count = copy.get(str);
                if (count == 1) copy.remove(str);
                else copy.put(str, count - 1);
                if (copy.isEmpty()) { // matches
                    res.add(i);
                    break;
                }
            } else break; // not in L
        }
    }
    return res;
}

 5 月 1 号

9  32 Longest Valid Parentheses  维护堆和一个开始

    public int longestValidParentheses(String s) {
        Stack<Integer> stack = new Stack<>();
        int start = 0, max = 0;
        for (int i = 0; i < s.length(); i++)
        {
            char c = s.charAt(i);
            if (c == '(')
            {
                stack.push(i);
            }
            else
            {
                if (stack.isEmpty())
                {
                    start = i + 1;
                }
                else
                {
                    stack.pop();
                    int curMax = stack.isEmpty() ? i - start + 1: i - stack.peek();
                    max = Math.max(max, curMax);
                }
            }
        }
        return max;
    }

10  38 Count and say  遍历 维护stringbuilder 

    public String countAndSay(int n) {
        if (n < 1) return "";
        String str = "1";
        for (int i = 2; i <= n; i++)
        {
            StringBuilder sb = new StringBuilder();
            int count = 1;
            for (int j = 1; j < str.length(); j++)
            {
                if (str.charAt(j) == str.charAt(j - 1))
                {
                    count++;
                }
                else
                {
                    sb.append(count);
                    sb.append(str.charAt(j - 1));
                    count = 1;
                }
            }
            sb.append(count);
            sb.append(str.charAt(str.length() - 1));
            str = sb.toString();
        }
        return str;
    }

11 49 Group Anagrams    维护hashmap

    public List<List<String>> groupAnagrams(String[] strs) {
        if (strs == null || strs.length == 0) return new ArrayList<List<String>>();
        HashMap<String, List<String>> map = new HashMap<>();
        for (int i = 0; i < strs.length; i++)
        {
            char[] c = strs[i].toCharArray();
            Arrays.sort(c);
            String cur = String.valueOf(c);
            if (!map.containsKey(cur)) map.put(cur, new ArrayList<String>());
            map.get(cur).add(strs[i]);
        }
        return new ArrayList<List<String>>(map.values());
    }

12 58 Length of Last Word   

    public int lengthOfLastWord(String s) {
        return s.trim().length() - s.trim().lastIndexOf(" ") - 1;
    }

 5 月 2 号

13  71 Simple path  双端队列

    public String simplifyPath(String path) {
        Deque<String> stack = new LinkedList<>();
        HashSet<String> set = new HashSet<>(Arrays.asList("..",".",""));
        for (String str : path.split("/")){
        if (str.equals("..") && !stack.isEmpty()) stack.pop();
        else if (!set.contains(str)) stack.push(str);
        }
        String res = "";
        for (String dir : stack) res = "/" + dir + res;
        return res.isEmpty()? "/" : res;
    }

14  91  Decode Ways     斐波纳数列

    public int numDecodings(String s) {
        if (s.isEmpty() || (s.length() > 1 && s.charAt(0) == '0')) return 0;
        int[] dp = new int[s.length() + 1];
        dp[0] = 1;
        for (int i = 1; i < dp.length; i++)
        {
            dp[i] = s.charAt(i - 1) == '0'? 0 : dp[i - 1];
            if (i > 1 && (s.charAt(i - 2) == '1' || (s.charAt(i - 2) == '2'&& s.charAt(i - 1) <= '6')))
            {
                dp[i] += dp[i - 2];
            }
        }
        return dp[s.length()];
    }

    public int numDecodings(String s) {
        if (s.isEmpty() || s.charAt(0) == '0') return 0;
         int c1 = 1, c2 = 1;
         for (int i = 1; i < s.length(); i++)
         {
             if (s.charAt(i) == '0') c1 = 0;
             if (s.charAt(i - 1) == '1' || s.charAt(i - 1) == '2' && s.charAt(i) <= '6')
             {
                 c1 = c1 + c2;
                 c2 = c1 - c2;
             }
             else
             {
                 c2 = c1;
             }
         }
         return c1;
    }

15 93 Restore  IP Addresses   4个循环

    public List<String> restoreIpAddresses(String s) {
    List<String> list = new ArrayList<>();
    for (int i = 1; i < 4 && i < s.length() - 2; i++)
    {
        for (int j = i + 1; j < i + 4 &&j < s.length() - 1; j++)
        {
            for (int k = j + 1; k < j + 4 && k < s.length(); k++)
            {
                String s1 = s.substring(0,i), s2 = s.substring(i,j), s3 = s.substring(j,k), s4 = s.substring(k,s.length());
                if (is(s1) && is(s2) && is(s3) && is(s4))
                {
                    list.add(s1 +"."+ s2 + "."+s3 + "."+s4);
                }
            }
        } 
    }
    return list;
    }
        public boolean is(String s){
        if(s.length()>3 || s.length()==0 || (s.charAt(0)=='0' && s.length()>1) || Integer.parseInt(s)>255)
            return false;
        return true;
    }

16  97  Interleaving String 动态规划

    public boolean isInterleave(String s1, String s2, String s3) {
        if (s1.length() + s2.length() != s3.length()) return false;
        boolean[][] res = new boolean[s1.length() + 1][s2.length() + 1];
        for (int i = 0; i < s1.length() + 1; i++)
        {
            for (int j = 0; j < s2.length() + 1; j++)
            {
                if (i == 0 && j == 0)
                {
                    res[i][j] = true;
                }
                else if (i == 0)
                {
                    res[i][j] = res[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
                }
                else if (j == 0)
                {
                    res[i][j] = res[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
                }
                else 
                {
            res[i][j] = res[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) || res[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
                }
            }
        }
        return res[s1.length()][s2.length()];
    }

 5 月 3 号

17  115 Distinct Subsequences

    public int numDistinct(String s, String t) {
        int[][] res = new int[t.length() + 1][s.length() + 1];
        for (int i = 0; i < s.length() + 1; i++)
        {
            res[0][i] = 1;
        }
        for (int i = 0; i < t.length(); i++)
        {
            for (int j = 0; j < s.length(); j++)
            {
                if (t.charAt(i) != s.charAt(j))
                {
                    res[i + 1][j + 1] = res[i + 1][j];
                }
                else
                {
                    res[i + 1][j + 1] = res[i + 1][j] + res[i][j];
                }
            }
        }
        return res[t.length()][s.length()];
    }

18 151  Reverse Words in a String    切分，反转，加入

    public String reverseWords(String s) {
        String[] words = s.trim().split(" +");
        Collections.reverse(Arrays.asList(words));
        return String.join(" ", words);
    }

19 157 Read N Characters Given read4

public class Solution extends Reader4 {
    public int read(char[] buf, int n) {
        for(int i = 0; i < n; i += 4){
            char[] tmp = new char[4];
            // 将数据读入临时数组
            int len = read4(tmp);
            // 将临时数组拷贝至buf数组，这里拷贝的长度是本次读到的个数和剩余所需个数中较小的
            System.arraycopy(tmp, 0, buf, i, Math.min(len, n - i));
            // 如果读不满4个，说明已经读完了，返回总所需长度和目前已经读到的长度的较小的
            if(len < 4) return Math.min(i + len, n);
        }
        // 如果循环内没有返回，说明读取的字符是4的倍数
        return n;
    }
}

20 158  Read N Characters Given read4 II

public class Solution extends Reader4 {
    Queue<Character> remain = new LinkedList<Character>();
    
    public int read(char[] buf, int n) {
        int i = 0;
        // 队列不为空时，先读取队列中的暂存字符
        while(i < n && !remain.isEmpty()){
            buf[i] = remain.poll();
            i++;
        }
        for(; i < n; i += 4){
            char[] tmp = new char[4];
            int len = read4(tmp);
            // 如果读到字符多于我们需要的字符，需要暂存这些多余字符
            if(len > n - i){
                System.arraycopy(tmp, 0, buf, i, n - i);
                // 把多余的字符存入队列中
                for(int j = n - i; j < len; j++){
                    remain.offer(tmp[j]);
                }
            // 如果读到的字符少于我们需要的字符，直接拷贝
            } else {
                System.arraycopy(tmp, 0, buf, i, len);
            }
            // 同样的，如果读不满4个，说明数据已经读完，返回总所需长度和目前已经读到的长度的较小的
            if(len < 4) return Math.min(i + len, n);
        }
        // 如果到这里，说明都是完美读取，直接返回n
        return n;
    }
}

5月 4 号

21  161 one edit distance  遍历，找到相同字符

    public boolean isOneEditDistance(string s, string t) { 
        int m = s.length(), n = t.length();
        for (int i = 0; i < Math.min(m, n); i++)
        {
            if (s.charAt(i) != t.charAt(i))
            {
                if (m == n) return s.substring(i + 1).equals(t.substring(i + 1));
                else if (m > n) return s.substring(i + 1).equals(t.substring(i));
                else return s.substring(i).equals(t.substring(i + 1));
            }
        }
        return Math.abs(m - n) == 1;
    }

22 165 Compare Version Numbers  切分，比较

    public int compareVersion(String version1, String version2) {
        String[] l1 = version1.split("\.");
        String[] l2 = version2.split("\.");
        int len = Math.max(l1.length, l2.length);
        for (int i = 0; i < len; i++)
        {
            Integer v1 = i < l1.length ? Integer.parseInt(l1[i]) : 0;
            Integer v2 = i < l2.length ? Integer.parseInt(l2[i]) : 0;
            int com = v1.compareTo(v2);
            if (com != 0) return com;
        }
        return 0;
    }

23  186 Reverse words in a string iI    反转全部，一个一个反转

 1 public void reverseWords(char[] s) {
 2     // Three step to reverse
 3     // 1, reverse the whole sentence
 4     reverse(s, 0, s.length - 1);
 5     // 2, reverse each word
 6     int start = 0;
 7     int end = -1;
 8     for (int i = 0; i < s.length; i++) {
 9         if (s[i] == ' ') {
10             reverse(s, start, i - 1);
11             start = i + 1;
12         }
13     }
14     // 3, reverse the last word, if there is only one word this will solve the corner case
15     reverse(s, start, s.length - 1);
16 }
17 
18 public void reverse(char[] s, int start, int end) {
19     while (start < end) {
20         char temp = s[start];
21         s[start] = s[end];
22         s[end] = temp;
23         start++;
24         end--;
25     }
26 }

24  214 Shortest Palindrome    end i j  找从开头最短

    public String shortestPalindrome(String s) {
     char[] c = s.toCharArray();
     int i = 0, j = c.length - 1, end = j;
     while (i < j)
     {
         if (c[i] == c[j])
         {
             i++;j--;
         }
         else
         {
             i = 0; end--; j = end;
         }
     }
     return new StringBuilder(s.substring(end + 1)).reverse().toString() + s;
    }

 5月5号

25 227 Basic Calculator II 栈  遇到符号加入

    int len;
    if(s==null || (len = s.length())==0) return 0;
    Stack<Integer> stack = new Stack<Integer>();
    int num = 0;
    char sign = '+';
    for(int i=0;i<len;i++){
        if(Character.isDigit(s.charAt(i))){
            num = num*10+s.charAt(i)-'0';
        }
        if((!Character.isDigit(s.charAt(i)) &&' '!=s.charAt(i)) || i==len-1){
            if(sign=='-'){
                stack.push(-num);
            }
            if(sign=='+'){
                stack.push(num);
            }
            if(sign=='*'){
                stack.push(stack.pop()*num);
            }
            if(sign=='/'){
                stack.push(stack.pop()/num);
            }
            sign = s.charAt(i);
            num = 0;
        }
    }

    int re = 0;
    for(int i:stack){
        re += i;
    }
    return re;

26 249 group shifted string (a[j] - a[0] + 26)%26

    public List<List<String>> groupStrings(String[] strings) {  
        List<List<String>> result = new ArrayList<List<String>>();  
        HashMap<String, List<String>> d = new HashMap<>();  
        for(int i = 0; i < strings.length; i++) {  
            StringBuffer sb = new StringBuffer();  
            for(int j = 0; j < strings[i].length(); j++) {  
                sb.append(Integer.toString(((strings[i].charAt(j) - strings[i].charAt(0)) + 26) % 26));  
                sb.append(" ");  
            }  
            String shift = sb.toString();  
            if(d.containsKey(shift)) {  
                d.get(shift).add(strings[i]);  
            } else {  
                List<String> l = new ArrayList<>();  
                l.add(strings[i]);  
                d.put(shift, l);  
            }  
        }  
          
        for(String s : d.keySet()) {  
            Collections.sort(d.get(s));  
            result.add(d.get(s));  
        }   
        return result;  
    }