414. Third Maximum Number

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
 

Example 3:

Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

Solution in C++：

關鍵點：

• 情況分類 && 初始值設定 && 資料更新

思路：

• 這裡主要知道題目的幾個邊界情況就行了，這裡主要的工作就是去重，即第一大的值和第二大的值不相同，依次類推。這裡對所有變數的初始值需要設定的比INT_MIN更小，以免輸入為INT_MIN時，最後判斷出錯；所以都採取範圍更大的long或者long long來進行。接下來就是對第三大的幾種分類及資料更新問題，記得將最大值修改後其後比它小的數也要跟著更新即可

自己程式碼

int thirdMax(vector<int>& nums) {

        size_t size = nums.size();

        if (size == 0)
            return 0;
        else if (size == 1)
            return nums[0];

        int max = nums[0];
        long long semax = LONG_MIN;
        long long thmax = LONG_MIN;
        bool flag = false;

        for(auto num : nums){
            
                
            if (num < max && num < semax && thmax < num){
                    flag = true;
                    thmax = num;
            } else if(num > max){
                if (semax < max){
                    if (thmax < semax){
                        flag = true;
                        thmax = semax;
                    }
                    semax = max;
                }
                max = num;
            } else if (num < max && num > semax){
                
                if (thmax < semax){
                    flag = true;
                    thmax = semax;
                }
                
                semax = num;
            }
        }

        if (!flag)
            return max;
        else
            return thmax;
    }
 

簡介版

int thirdMax(vector<int>& nums) {
    if (nums.size() == 0) {
			return 0;
		}
		long firstMax = LONG_MIN;
		long secondMax = LONG_MIN;
		long thirdMax = LONG_MIN;
		for (int num : nums) {
			if (num > firstMax) {
				thirdMax = secondMax;
				secondMax = firstMax;
				firstMax = num;
			} else if (num > secondMax && num < firstMax) {
				thirdMax = secondMax;
				secondMax = num;
			} else if (num > thirdMax && num < firstMax && num < secondMax) {
				thirdMax = num;
			}
		}
		return (int) (thirdMax == LONG_MIN ? firstMax : thirdMax);
    }

415. Add Strings

Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.

Note:

1. The length of both num1 and num2is < 5100.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

Solution in C++：

關鍵點：

• 加法進位

思路：

• 就是簡單的加法操作，根據字串長度不同等不同情況進行處理即可。主要比較繁瑣的點在於長度不一致及進位處理問題，但是這個彷彿在高手眼中不算事兒啊，見簡潔版程式碼。

自己程式碼

string addStrings(string num1, string num2) {
        string result;
        size_t size1 = num1.size();
        size_t size2 = num2.size();
        size_t minsize = size1 < size2 ? size1 : size2;
        int carry = 0;
        
        for(int i = 1; i <= minsize; ++i){
            int tmp = num1[size1 - i] + num2[size2 -i] - 2 * '0' + carry;
            if (tmp > 9){
                carry = 1;
                tmp = tmp - 10;
            }
            else
                carry = 0;
            result = to_string(tmp) + result;
        }
        
        if (carry == 0){ // 無進位
            if (size1 > size2)
                result = num1.substr(0, size1 - size2) + result;
            else 
                result = num2.substr(0, size2 - size1) + result;
        } else{         // 有進位
            if (size1 > size2){
                 for(int i = size1 - size2 - 1; i >=0 ; --i){
                     int tmp = num1[i] -'0' + carry;
                     if (tmp > 9){
                        carry = 1;
                        tmp = tmp - 10;
                    }
                    else
                        carry = 0;
                    result = to_string(tmp) + result;
                 }
                     
            } else if (size1 < size2){
                for(int i = size2 - size1 - 1; i >=0 ; --i){
                     int tmp = num2[i] - '0' + carry;
                     if (tmp > 9){
                        carry = 1;
                        tmp = tmp - 10;
                    }
                    else
                        carry = 0;
                    result = to_string(tmp) + result;
                 }
            }          
        }
        
        if (carry == 1)
            result = to_string(1) + result;
        
        return result;
    }

簡潔版

string addStrings(string num1, string num2) {
        string result;
        int carry = 0;
        for(int i = num1.size() - 1, j = num2.size() - 1; i >= 0 || j >= 0 || carry == 1; i--, j--){
            int x = i < 0 ? 0 : num1[i] - '0';
            int y = j < 0 ? 0 : num2[j] - '0';
            result = to_string((x + y + carry) % 10) + result;
            carry = (x + y + carry) / 10;
        }
        return result;
    }

434. Number of Segments in a String

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters.

Please note that the string does not contain any non-printable characters.

Example:

Input: "Hello, my name is John"
Output: 5

Solution in C++：

關鍵點：

• 頭尾及連續空格處理

思路：

• 其實我開始想的也是簡潔版的這種做法，但是鑑於下標越不越界的控制我有點糊，然後就換了思維了，去連續的空格。然後發現大佬就是大佬啊，這種輕而易舉去掉第一個的做法也是漲知識了。

自己程式碼

int countSegments(string s) {
        int count = 0;
        size_t size = s.size();
        
        // 去掉頭部空格
        int i = 0;
        while(s[i] == ' ')
            ++i;
        
        // 去掉尾部空格
        int j = 0;
        while(s[size - j - 1] == ' ')
            ++j;
        
        bool flag = false;
        for(; i < size - j; ++i){
            flag = true;
            if (s[i] == ' '){
                ++count;
                while(s[i] == ' ')
                    ++i;
                --i;
            }      
        }
        
        if (flag)
            return count+1;
        else
            return count;
    }

簡潔版

int countSegments(string s) {
        int count = 0;
        size_t size = s.size();
        for(int i = 0; i < size; ++i){
            if ((i == 0 || s[i-1] == ' ') && s[i] != ' ')
                ++count;
        }
        return count;
    }

438. Find All Anagrams in a String

Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

Solution in C++：

關鍵點：

• 哈哈

思路：

• 很暴力的方法，就是遍歷，然後判斷是不是p的anagrams，這個判斷之前刷的題裡面也有過，大致思想就是字母數量一致即可。

vector<int> findAnagrams(string s, string p) {
        vector<int> result;
        size_t sizes = s.size();
        size_t sizep = p.size();
        
        if (sizes < sizep || sizes == 0)
            return result;
        
        for(int i = 0; i < sizes - sizep + 1; ++i){
            // 判斷是否s[i , i+sizep]為p的anagrams
            bool flag = true;
            vector<int> letters(26,0);
            for(int j = 0; j < sizep; ++j){
                ++letters[p[j]-'a'];
                --letters[s[i+j]-'a'];
            }
            for(auto letter : letters)
                if(letter != 0)
                    flag = false;
            if (flag)
                result.push_back(i);
        }
        
        return result;
    }

小結

哇，今天刷題還不錯，昨天的程式碼就改了範圍就跑通了，然後就是接下來的兩題，寫的程式碼看都不想看，不知道想的有多複雜，哎，可能還是分情況討論能力不太強，看到別人的簡潔程式碼我真是羨慕不已啊。

知識點

• clean code
• 分情況