阿华代码，不是逆风，就是我疯

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目录

 一：替换所有的问号

二：提莫攻击

三：z字形变换

四：外观数列

五：数青蛙

方法一

方法二

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 一：替换所有的问号

1576. 替换所有的问号 - 力扣（LeetCode）

方法一

class Solution {public String modifyString(String ss) {char[] s = ss.toCharArray();int len = s.length;for(int i = 0 ; i < s.length ; i++){if(s[i] == '?'){for(char ch = 'a' ; ch <= 'z' ; ch++){if( (i == 0 || ch != s[i-1]) && (i == len-1 || ch != s[i+1]) ){s[i] = ch;break;}}}}String ret = String.valueOf(s);return ret;}
}

方法二（复杂） 

class Solution {public String modifyString(String s) {char[] tem = new char[]{'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};char[] str = s.toCharArray();for(int i = 0 ; i < str.length ; i++){while(str[i] == '?'){for(int j = 0 ; j < tem.length ; j++){if(i == 0){if((str.length > 1) && (str[i+1] != tem[j])){str[i] = tem[j];break;}else if(str.length == 1){str[i] = tem[j];break;}else{}}else if(i == str.length-1){if(str[i-1] != tem[j]){str[i] = tem[j];}else{}}else{if((str[i+1] != tem[j]) && (str[i-1] != tem[j])){str[i] = tem[j];break;}else{}}}}}String ret = "";for(int i = 0 ; i < str.length ; i++){ret += str[i];}return ret;}
}

二：提莫攻击

495. 提莫攻击

class Solution {public int findPoisonedDuration(int[] nums, int time) {int sum = 0;for(int i = 0 ; i < nums.length-1 ; i++){if(nums[i+1] - nums[i] >= time){sum += time; }else{sum += nums[i+1]-nums[i];}}sum += time;return sum;}
}

三：z字形变换

6. Z 字形变换

class Solution {public String convert(String s, int numRows) {//处理一下特殊情况if(numRows == 1){return s;}//1：创建一个变量来接收拼接的字符串int d = 2 * numRows - 2;StringBuilder ret = new StringBuilder();int len = s.length();//2:确定第一行的下标，并按照下标找到该字符，拼接上去for(int i = 0 ; i < len ; i += d){ret.append(s.charAt(i));}//3：确定中间的行数，两层循环，一行一行处理，需要两个指针用于确定坐标，然后拼接for(int i = 1 ; i < numRows-1 ; i++){for(int k = i , j = d - k ; k < len || j < len ; k += d , j += d){if(k < len){ret.append(s.charAt(k));}if(j < len){ret.append(s.charAt(j));}}}//4：确定好最后一行，同第一行处理方式一样for(int i = numRows-1 ; i < len ; i += d){ret.append(s.charAt(i));}return ret.toString();}
}

四：外观数列

38. 外观数列

class Solution {public String countAndSay(int n) {String ret = "1";//1:下面的循环用于依次解释ret,只用解释n-1次for(int i = 1 ; i <= n-1 ; i++){//每一次都是一个新的StringBuilderStringBuilder temp = new StringBuilder();//模拟：第一次解释，对应的是n=2int len = ret.length();for(int left = 0 , right = 0 ; right < len ; ){//2:模拟有几个相同的数while(right < len && ret.charAt(left) == ret.charAt(right)){right++;}temp.append(Integer.toString(right-left));temp.append(ret.charAt(left));left = right;}//更新ret,StringBuilder转换为String类型ret = temp.toString();}return ret;}
}

五：数青蛙

1419. 数青蛙

方法一

class Solution {public int minNumberOfFrogs(String croakOfFrogs) {char[] str = croakOfFrogs.toCharArray();String t = "croak";int n = t.length();int[] hash = new int[n];Map<Character,Integer> hashMap = new HashMap<>();//<字符，字符对应下标>for(int i = 0 ; i < n ; i++){hashMap.put(t.charAt(i),i);}for(char ch : str){if(ch == t.charAt(0)){if(hash[n-1] != 0){hash[n-1]--;hash[0]++;}else{hash[0]++;}}else{int i = hashMap.get(ch);if(hash[i-1] == 0){return -1;}else{hash[i-1]--;hash[i]++;}}}for(int i = 0 ; i < n-1 ; i++){if(hash[i] != 0){return -1;}}return hash[n-1];}
}

方法二

class Solution5 {public int minNumberOfFrogs(String croakOfFrogs) {//创建hash表，创建字符数组，遍历数组Map<Character,Integer> hash = new HashMap<>();char[] str = croakOfFrogs.toCharArray();hash.put('c',0);hash.put('r',0);hash.put('o',0);hash.put('a',0);hash.put('k',0);for(int i = 0 ; i < str.length ; i++){if(str[i] == 'c'){if(hash.get('k') == 0){hash.put('c',hash.get('c')+1);}else{hash.put('c',hash.get('c')+1);hash.put('k',hash.get('k')-1);}}else{if(str[i] == 'r'){if(hash.get('c') != 0){hash.put('c',hash.get('c')-1);hash.put('r',hash.get('r')+1);}else{return -1;}}else if(str[i] == 'o'){if(hash.get('r') != 0){hash.put('r',hash.get('r')-1);hash.put('o',hash.get('o')+1);}else{return -1;}}else if(str[i] == 'a'){if(hash.get('o') != 0){hash.put('o',hash.get('o')-1);hash.put('a',hash.get('a')+1);}else{return -1;}}else if(str[i] == 'k'){if(hash.get('a') != 0){hash.put('a',hash.get('a')-1);hash.put('k',hash.get('k')+1);}else{return -1;}}else{}}}if(hash.get('c') != 0 || hash.get('r') != 0 ||hash.get('o') != 0 || hash.get('a') != 0 ){return -1;}int ret = hash.get('k');return ret;//判断}
}