【LeetCode】水题（刚开始重新刷题找感觉用的）

[9] Palindrome Number [Easy]

给一个数字，用不转化成字符串的方式判断它是否是回文。

先求数字长度，然后把数字的后半段做翻转(就是不断地取模，除10这种方式)，然后判断前后半段是否相等

有特殊情况，负数，10的倍数，和零

 1 class Solution {
 2 public:
 3     bool isPalindrome(int x) {
 4         if (x < 0 || x % 10 == 0 && x != 0) {
 5             return false;
 6         }
 7         //=============[1]先求数字长度
 8         int len = 1;
 9         int tmpx = x;
10         while (tmpx / 10 != 0) {
11             tmpx /= 10;
12             len++;
13         }
14         cout << len << endl;
15         
16         //=============[2]然后翻转后面一半数字
17         int reverse = 0;
18         tmpx = x;
19         for (int i = 1; i < len / 2 + 1; ++i) {
20             reverse = reverse * 10 + (tmpx % 10);
21             tmpx /= 10;
22         }
23         cout << tmpx << "   " << reverse << endl;
24         
25         //=============[3]根据长度的奇偶性判断是否需要再除以10
26         if (len % 2 == 1) {
27             tmpx /= 10;
28         } 
29         if (tmpx != reverse) {
30             return false;
31         }
32         return true;
33     }
34 };

View Code

[157] Read N Characters Given Read4 [Easy]

给个read4的api，每次能从文件里面读四个字符，最多读N个字符，考虑到文件里面的字符数目可能比N小，所以不能直接返回N。

每次应该加上 min(read4(buf + k), n); 这种表达，代码如下：

 1 // Forward declaration of the read4 API.
 2 int read4(char *buf);
 3 
 4 class Solution {
 5 public:
 6     /**
 7      * @param buf Destination buffer
 8      * @param n   Maximum number of characters to read
 9      * @return    The number of characters read
10      */
11     int read(char *buf, int n) {
12         int res = 0, k = 0;
13         while (n > 0) {
14             int readN = min(read4(buf + k*4), n);
15             n = n - 4;
16             ++k;
17             res += readN;
18         }
19         return res;
20     }
21 };

View Code

[168] Excel Sheet Column Title [Easy]

给个数字N，返回对应excel上面的列名。比如1返回A，2返回B，27返回AA

26进制换算，有个N--原因需要注意。

 1 class Solution {
 2 public:
 3     string convertToTitle(int n) {
 4         string ans = "";
 5         while(n > 0) {
 6             n--; //在这里N--的原因是A对应1，Z对应26
 7             unsigned int  mod = n % 26;
 8             char  ch = 'A' + mod;
 9             ans = ch + ans;
10             n = n / 26;
11         }
12         return ans ;
13     }
14 };

View Code

[266] Palindrome Permutation [Easy]

给个字符串，判断用它的字母能不能组成个回文串，秒杀

 1 class Solution {
 2 public:
 3     bool canPermutePalindrome(string s) {
 4         map<char, int> mapCnt;
 5         for (auto ele : s) {
 6             mapCnt[ele]++;
 7         }
 8         int oddCnt = 0;
 9         for (auto ele : mapCnt) {
10             if(ele.second % 2) {
11                 oddCnt++;   
12                 if (oddCnt >= 2) {
13                     return false;
14                 }
15             }
16         }
17         return true;
18     }
19 };

View Code

[500] Keyboard Row [Easy]

给N个单词，判断哪些单词的字母在键盘的同一行，输出这些单词。

 1 class Solution {
 2 public:
 3     set<char> setLine1{'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'}, 
 4             setLine2{'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'},
 5             setLine3{'z', 'x', 'c', 'v', 'b', 'n', 'm'};
 6 
 7     int findGroup(char lowerChar) {
 8         if (setLine1.find(lowerChar) != setLine1.end()) {
 9             return 1;
10         } else if (setLine2.find(lowerChar) != setLine2.end()) {
11             return 2;
12         } else if (setLine3.find(lowerChar) != setLine3.end()){
13             return 3;
14         }
15         return -1;
16     }
17     
18     vector<string> findWords(vector<string>& words) {
19         vector<string> answer;
20         for(auto word : words) {
21             int groupId = 0;
22             bool ansIn = true;
23             groupId = isupper(word[0]) ? findGroup(tolower(word[0])) : findGroup(word[0]);
24             for (auto i = 1; i < word.size(); ++i) {
25                 int tmpGroupId = 0;
26                 tmpGroupId = isupper(word[i]) ? findGroup(tolower(word[i])) : findGroup(word[i]);
27                 if (tmpGroupId != groupId) {
28                     ansIn = false;
29                     break;
30                 }
31             }
32             if (ansIn) {
33                 answer.push_back(word);
34             }
35         }
36         return answer;
37     }
38 };

View Code