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'};
HashMap<Character, Integer> hash = new HashMap<>();
HashMap<Character, 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";
HashMap<Character, Integer> hashIndex = new HashMap<>();
for (int i = 0; i < 5; i++) {
hashIndex.put(str.charAt(i), i);
}
HashMap<Character, 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,k
char 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 {
// c
if (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];
HashMap<Character, 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,k
int i = index.get(ch);
if (hash[i - 1] > 0) {
hash[i - 1]--;
hash[i]++;
} else {
return -1;
}
} else {
// c
if (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~
