郑州网站建设模板制作,企业网站报价模板下载,深圳宝安网站建设,wordpress琪亚娜文章目录 模拟替换所有的问号提莫攻击Z 字形变换外观数列数青蛙 总结 模拟
替换所有的问号 按照题目的要求写代码即可~ public String modifyString(String ss) {int n ss.length();if (n 1) {return a;}char[] s ss.toCharArray();for (int i 0; i n; i… 文章目录 模拟替换所有的问号提莫攻击Z 字形变换外观数列数青蛙 总结 模拟
替换所有的问号 按照题目的要求写代码即可~ public String modifyString(String ss) {int n ss.length();if (n 1) {return a;}char[] s ss.toCharArray();for (int i 0; i n; i) {if (s[i] ?) {for (char ch a; ch z; ch) {if (i 0 ch ! s[i 1]) {// 第一个s[i] ch;} else if (i n - 1 ch ! s[i - 1]) {// 最后一个s[i] ch;}if (0 i i n - 1 ch ! s[i 1] ch ! s[i - 1]) {// 中间s[i] ch;}}}}return String.valueOf(s);}题解写的更加简洁.
题解代码: public String modifyString(String ss) {int n ss.length();char[] s ss.toCharArray();for (int i 0; i n; i) {if (s[i] ?) {for (char ch a; ch z; ch) {if ((i 0 || s[i - 1] ! ch) (i n - 1 || s[i 1] ! ch)) {s[i] ch;break;}}}}return String.valueOf(s);}提莫攻击 草稿: public int findPoisonedDuration(int[] timeSeries, int duration) {int tmp timeSeries[0] duration - 1;int sum duration;for (int i 1; i timeSeries.length; i) {if (tmp timeSeries[i]) {sum timeSeries[i] - timeSeries[i - 1];} else {sum duration;}tmp timeSeries[i] duration - 1;}return sum;}题解代码: 草图: public int findPoisonedDuration(int[] timeSeries, int duration) {int sum 0;for (int i 1; i timeSeries.length; i) {int tmp timeSeries[i] - timeSeries[i - 1];if (tmp duration) {sum duration;} else {sum tmp;}}return sum duration;}Z 字形变换 虽然过了,但是稀里糊涂地过了~
开头和结尾都好说,主要是中间,不知道为啥要 - 2*i. 规律就是这样的~ 做题思路就是:
题目让干啥,我们就干啥画图找规律~
坑:
numRows 可能为 1 .放中间元素时,容易越界.
代码:
public String convert(String ss, int numRows) {if (numRows 1)return ss;char[] s ss.toCharArray();int n s.length;char[] ret new char[n];int gap (numRows - 1) * 2;int k 0;// 开头for (int j 0; j n; j gap) {ret[k] s[j];}// 中间for (int i 1; i numRows - 2; i) {for (int j i; j n; j gap) {ret[k] s[j];// 这里为啥 - i*2 就对了?int mid j gap - i * 2;if (mid n) {ret[k] s[mid];}}}// 结尾for (int j numRows - 1; j n; j gap) {ret[k] s[j];}return String.valueOf(ret);}外观数列 终于过了~ 不知道为啥,自己写的代码返回的结果一直只有两个数. 在这上面耗了20多分钟. 最后全删了.心态崩了呀. 吃完饭回来,重写了一遍,只用了不到6分钟就写出来了.
坑:
不用考虑怎么替换的问题,最开始我也被题目带偏了.如果用替换来写,需要考虑的情况就复杂了. 其实直接新建一个字符串,不断向这个字符串后面拼接就行了. public String countAndSay(int n) {StringBuilder ret new StringBuilder(1);for (int i 1; i n; i) {StringBuilder tmp new StringBuilder();int len ret.length();int left 0, right 0;while (right len) {while (right len ret.charAt(left) ret.charAt(right)) {right;}tmp.append(right - left);tmp.append(ret.charAt(left));left right;}ret tmp;}return ret.toString();}数青蛙 最后一个测试用例卡了好久.
坑:
如何判断给出的字符串不是 “croak” 的有效组合? 可以用最后的 sum 来判断,如果 sum 没有减到0,那就说明字符串不完整. public int minNumberOfFrogs(String croakOfFrogs) {if (croakOfFrogs.length() 5 || croakOfFrogs.length() % 5 ! 0) {return -1;}int sum 0;int ret 0;char[] str {c, r, o, a, k};HashMapCharacter, Integer hash new HashMap();HashMapCharacter, Character hash2 new HashMap();for (int i 1; i 5; i) {hash2.put(str[i], str[i - 1]);}int n croakOfFrogs.length();for (int i 0; i n; i) {char ch croakOfFrogs.charAt(i);hash.put(ch, hash.getOrDefault(ch, 0) 1);if (ch ! c hash.getOrDefault(ch, 0) hash.getOrDefault(hash2.get(ch), 0)) {return -1;}if (ch c) {sum;} else if (ch k) {ret Math.max(ret, sum);sum--;}}if (sum ! 0) return -1;return ret;}看了题解后又自己写了一遍: public int minNumberOfFrogs(String croakOfFrogs) {String str croak;HashMapCharacter, Integer hashIndex new HashMap();for (int i 0; i 5; i) {hashIndex.put(str.charAt(i), i);}HashMapCharacter, Integer hashCount new HashMap();int n croakOfFrogs.length();for (int i 0; i n; i) {char ch croakOfFrogs.charAt(i);if (ch ! c) {// r,o,a,kchar prev str.charAt(hashIndex.get(ch) - 1);int pervCount hashCount.getOrDefault(prev, 0);if (pervCount 0) {hashCount.put(prev, pervCount - 1);hashCount.put(ch, hashCount.getOrDefault(ch, 0) 1);} else if (pervCount 0) {return -1;}} else {// cif (hashCount.getOrDefault(k, 0) 0) {hashCount.put(k, hashCount.get(k) - 1);}hashCount.put(ch, hashCount.getOrDefault(ch, 0) 1);}}// 检验给出的字符串是不是 croak 的有效组合。for (int i 0; i 4; i) {if (hashCount.get(str.charAt(i)) ! 0) {return -1;}}return hashCount.get(k);}题解代码:
使用数组替代了 hash 表. public int minNumberOfFrogs(String c) {char[] croakOfFrogs c.toCharArray();String str croak;int n str.length();int[] hash new int[n];HashMapCharacter, Integer index new HashMap();// 建立字母和下标的关系for (int i 0; i n; i) {index.put(str.charAt(i), i);}for (char ch : croakOfFrogs) {if (ch ! c) {// r,o,a,kint i index.get(ch);if (hash[i - 1] 0) {hash[i - 1]--;hash[i];} else {return -1;}} else {// cif (hash[n - 1] 0)hash[n - 1]--;hash[0];}}for (int i 0; i n - 1; i) {if (hash[i] ! 0) return -1;}return hash[n - 1];}看题解看到了一个 if else 大法 :
public int minNumberOfFrogs(String croakOfFrogs) {int c,r,o,a,k;c 0; r 0; o 0; a 0;k 0;char []chars croakOfFrogs.toCharArray();int res 0;for(int i 0;i chars.length;i){if(chars[i] c){if(k 0){k--;}else{res;}c;}else if(chars[i] r){c--;r;}else if(chars[i] o){r--;o;}else if(chars[i] a){o--;a;}else if(chars[i] k){a--;k;}if(c 0 || r 0 || o 0 || a 0){break;}}if(c ! 0 || r ! 0 || o ! 0 || a ! 0){return -1;}return res;}总结
做模拟题时, 题目说啥咱干啥~有难度的模拟题需要我们找规律.画图是个好东西. 本文到这里就结束啦~
