一、反转问题

1.反转字符串

思路：双指针，反转数组一个套路

class Solution {public void reverseString(char[] s) {int l = 0;int r = s.length -1;while(l<r){char c = s[l];s[l] = s[r];s[r] = c;l++;r--;}}
}

2.k个一组反转

思路：每k个进行一次反转函数，从i++变成i+=2*k

class Solution {public String reverseStr(String s, int k) {if(s == null || s.length() == 0){return s;}int len = s.length();char[] str = s.toCharArray();for(int i = 0;i<len;i+=2*k){revers(str,i,Math.min(i+k-1,len-1));}return new String(str);}public void revers(char[] s,int l,int r){while(l<r){char c = s[l];s[l] = s[r];s[r] = c;l++;r--;}}
}

3.仅仅反转字母

思路：

方法一、双指针

class Solution {public String reverseOnlyLetters(String s) {if(s==null || s.length() == 0){return s;}int l = 0;int r = s.length() -1;char[] str = s.toCharArray();while(l<r){while(l<str.length&&!Character.isLetter(str[l])){l++;}while(r>=0&&!Character.isLetter(str[r])){r--;}if(l<r){char c = str[l];str[l] = str[r];str[r] = c;l++;r--;}}return new String(str);}
}

方法二、使用栈，实现出栈反转

字母压入栈中，遇到字母弹出栈顶拼接进结果集

class Solution {public String reverseOnlyLetters(String s) {if(s==null || s.length() == 0){return s;}char[] str = s.toCharArray();Stack<Character> stack = new Stack();for(int i = 0;i<str.length;i++){if(Character.isLetter(str[i])){stack.push(str[i]);}}StringBuilder sb = new StringBuilder();for(char c : str){if(Character.isLetter(c)){sb.append(stack.pop());}else{sb.append(c);}}return new String(sb);}
}

4.反转字符串中的单词

思路：整体反转+局部反转 = 区间反转

class Solution {public String reverseWords(String s) {StringBuilder sb = trimSpaces(s);int len = sb.length();reverse(sb,0,len - 1);int i = 0;for(int j = 0;j<=len;j++){if(j == len||sb.charAt(j) == ' '){reverse(sb,i,j-1);i = j+1;}}return sb.toString();}public StringBuilder trimSpaces(String s) {int left = 0, right = s.length() - 1;// 去掉字符串开头的空白字符while (left <= right && s.charAt(left) == ' ') {++left;}// 去掉字符串末尾的空白字符while (left <= right && s.charAt(right) == ' ') {--right;}// 将字符串间多余的空白字符去除StringBuilder sb = new StringBuilder();while (left <= right) {char c = s.charAt(left);if (c != ' ') {sb.append(c);} else if (sb.charAt(sb.length() - 1) != ' ') {sb.append(c);}++left;}return sb;}public void reverse(StringBuilder sb, int left, int right) {while (left < right) {char tmp = sb.charAt(left);sb.setCharAt(left++, sb.charAt(right));sb.setCharAt(right--, tmp);}}}

二、验证回文串

思路：双指针，跳过非字母数字

class Solution {public boolean isPalindrome(String s) {if(s == null || s.length() == 0){return true;}int l = 0;int r = s.length() - 1;char[] str = s.toCharArray();while(l<r){while(l<r&&!Character.isLetterOrDigit(str[l])){l++;}while(l<r&&!Character.isLetterOrDigit(str[r])){r--;}if(Character.toLowerCase(str[l]) != Character.toLowerCase(str[r])){return false;}l++;r--;}return true;}
}

三、字符串中的第一个唯一字符

 思路：哈希表来映射索引，当出现次数超过1就赋值-1标记，最后遍历一遍哈希表返回最小值

class Solution {public int firstUniqChar(String s) {HashMap<Character,Integer> map = new HashMap<>();int min = Integer.MAX_VALUE;int len = s.length();for(int i = 0;i<len;i++){char c = s.charAt(i);if(map.containsKey(c)){map.put(c,-1);}else{map.put(c,i);}   }for(Object o : map.values()){Integer v = (Integer) o;if(v != -1 && v<min){min = v;}}return min == Integer.MAX_VALUE ? -1 : min;}
}

四、有效的字母异位词

思路：

方法一、排序，之后比较是否相等

class Solution {public boolean isAnagram(String s, String t) {if (s.length() != t.length()) {return false;}char[] str1 = s.toCharArray();char[] str2 = t.toCharArray();Arrays.sort(str1);Arrays.sort(str2);return Arrays.equals(str1, str2);}
}

方法二、哈希表存储，比较出现次数是否相同

数组实现

class Solution {public boolean isAnagram(String s, String t) {int[] temp = new int[26];for(char c : s.toCharArray()){temp[c-'a'] += 1; }for(char c : t.toCharArray()){temp[c-'a'] -= 1; }for(int i : temp){if(i != 0){return false;}}return true;}
}

集合实现，更有通用性，不是字母也可以比较

class Solution {public boolean isAnagram(String s, String t) {Map<Character,Integer> maps = new HashMap<>();Map<Character,Integer> mapt = new HashMap<>();for(Character c : s.toCharArray()){maps.put(c,maps.getOrDefault(c,0)+1);}for(Character c : t.toCharArray()){mapt.put(c,mapt.getOrDefault(c,0)+1);}for(Object key  : maps.keySet()){if(!mapt.containsKey(key) || !maps.get(key).equals(mapt.get(key))){return false;}mapt.remove(key);}return mapt.isEmpty() ? true : false;}
}