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); }
}

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