LEETCODE65字符分类，medium&easy级别，题目：206473

Posted 2022-06-11 cutter-point

 

今天的字符类还比较简单

package y2019.Algorithm.str.easy;

import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

/**
 * @ClassName IsValid
 * @Description 20. Valid Parentheses
 *
 * Given a string containing just the characters ‘(‘, ‘)‘, ‘‘, ‘‘, ‘[‘ and ‘]‘, determine if the input string is valid.
 *
 * An input string is valid if:
 *
 * Open brackets must be closed by the same type of brackets.
 * Open brackets must be closed in the correct order.
 * Note that an empty string is also considered valid.
 *
 * @Author xiaof
 * @Date 2019/8/4 15:46
 * @Version 1.0
 **/
public class IsValid 

    public boolean solution(String s) 

        Map comMap = new HashMap();
        Stack stack = new Stack();
        comMap.put(‘(‘, ‘)‘);comMap.put(‘[‘, ‘]‘);comMap.put(‘‘, ‘‘);
        //1.遍历字符串，获取每一个字符
        char cs[] = s.toCharArray();
        for (int i = 0; i < cs.length; ++i) 
            //2.判断是是否是：(,[,中的字符，如果是那么就入栈，如果不是就出栈
            if(comMap.containsKey(cs[i])) 
                //如果key包含
                stack.push(cs[i]);
             else 

                if(stack.size() <= 0) 
                    return false;
                

                //3.判断出栈的数据和当前的数据是否正好配对，如果是，那么就ok，如果不是，那么就false
                char temp = (char) stack.pop();
                if((char) comMap.get(temp) != cs[i]) 
                    //如果不等
                    return false;
                
            
        

        if(stack.size() > 0) 
            return false;
        

        return true;

    

    public static void main(String[] args) 
        String s = "()";

        IsValid fuc = new IsValid();

        fuc.solution(s);

    

 

package y2019.Algorithm.str.medium;

/**
 * @ClassName CountSubstrings
 * @Description 647. Palindromic Substrings
 *
 * Given a string, your task is to count how many palindromic substrings in this string.
 *
 * The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
 *
 * Example 1:
 *
 * Input: "abc"
 * Output: 3
 * Explanation: Three palindromic strings: "a", "b", "c".
 *
 *
 * Example 2:
 *
 * Input: "aaa"
 * Output: 6
 * Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
 *
 * 统计字符中是否包含回文字符的字串
 *
 * @Author xiaof
 * @Date 2019/8/4 16:30
 * @Version 1.0
 **/
public class CountSubstrings 

    public int solution(String s) 
        //双层循环遍历所有字串
        int count = 0;
        char[] source = s.toCharArray();
        for (int i = 0; i < source.length; ++i) 
            for (int j = i; j < source.length; ++j) 
                //遍历所有的字符串
                if(isPalindromic(i, j, source)) 
                    count++;
                
            
        

        return count;
    

    /**
     *
     * @param l 左边索引
     * @param r 右边索引
     * @param source 原始字符的字符数组
     * @return
     */
    public boolean isPalindromic(int l, int r, char[] source) 
        while (l <= r) 
            if (source[l] == source[r]) 
                ++l;
                --r;
             else 
                return false;
            
        

        return true;
    

package y2019.Algorithm.str.medium;

import java.util.HashMap;
import java.util.Map;

/**
 * @ClassName LengthOfLongestSubstring
 * @Description 3. Longest Substring Without Repeating Characters
 *
 * Given a string, find the length of the longest substring without repeating characters.
 *
 * Example 1:
 *
 * Input: "abcabcbb"
 * Output: 3
 * Explanation: The answer is "abc", with the length of 3.
 * Example 2:
 *
 * Input: "bbbbb"
 * Output: 1
 * Explanation: The answer is "b", with the length of 1.
 * Example 3:
 *
 * Input: "pwwkew"
 * Output: 3
 * Explanation: The answer is "wke", with the length of 3.
 *              Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
 *
 * @Author xiaof
 * @Date 2019/8/4 17:30
 * @Version 1.0
 **/
public class LengthOfLongestSubstring 

    public int solution(String s) 

        if(s == null || s.equals("")) 
            return 0;
        

        //统计最长连续子字串，那么我们只需要每次剔除重复的那个字符，然后从那个位置开始就可以了
        int start = 0, count = 1;
        Map num = new HashMap();
        boolean lastcount = false;
        char[] source = s.toCharArray();
        num.put(source[0], 0);

        for (int i = 1; i < source.length; ++i) 
            //判断前面的字串中是否有包含,这里要有(int) num.get(source[i]) >= start，而且是大于等于，避免之前跳过的数据干扰，并且不能排除掉起始位置
            if (num.containsKey(source[i]) && (int) num.get(source[i]) >= start) 
                //如果包含了，说明之前已经出现重复的字串，那么统计一波
                count = Math.max(count, (i - start));
                start = (int) num.get(source[i]) + 1;
                num.put(source[i], i);
             else 
                num.put(source[i], i);
                if (i == source.length - 1) 
                    lastcount = true;
                
            
        

        //循环到最后，计算最后一个位置
        if (lastcount) 
            count = Math.max(count, (source.length - start));

        

        return count;

    

    public static void main(String[] args) 
        String s = "pwwkew";
        String s1 = "tmmzuxt";
        String s2 = "abcabcbb";

        LengthOfLongestSubstring fuc = new LengthOfLongestSubstring();

        fuc.solution(s2);