LeetCode第一周总结

LeetCode 383. Ransom Note

https://leetcode.com/problems/ransom-note/ 原题地址

题目描述

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

解法：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> num(26);
        for(int i = 0; i < magazine.size(); i++)## 遍历 magazine，巧妙利用数组下标
            num[magazine[i] - 'a']++;
        for(int i = 0; i < ransomNote.size(); i++){
            num[ransomNote[i] - 'a']--;
            if(num[ransomNote[i]- 'a'] < 0)## 出现负数则返回false
                return false;
        }
            return true; 
    }
};

 LeetCode 344. Reverse String

https://leetcode.com/problems/reverse-string/原题地址

Write a function that reverses a string. The input string is given as an array of characters char[].

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

You may assume all the characters consist of printable ascii characters.

 

Example 1:

Input: ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]

Example 2:

Input: ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

解法：
　　简单的申请一个char类型的字符，然后遍历就可以
　　运行时间 > 43%，效率一般

class Solution {
public:
    void reverseString(vector<char>& s) {
        for(int i = 0; i < s.size()/2; i++){
            char c = s[i];
            s[i] = s[s.size() -i -1];
            s[s.size() -i -1] = c;
        }
            
    }
};#52ms faster than 43%

LeetCode 67. Add Binary

题目描述：

　　

Given two binary strings, return their sum (also a binary string).

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

解法：
   1、统计两个字符串的长度，
   2、如果字符串先遍历到首位则补零

   3、从右向左逐位相加（并加上进位carry）

   4、计算当前位和进位值
   5、最后判断进位值是否为1，

　　例：   string1 = "1101";   string2 = "111"变为->"0111"

　　确定进位carry的值，因为是二进制 sum % 2 为当前位置的值，sum / 2 为进位值
    3/2 == 1进位1，2/2进位1，1/2进位0，0/2进位0
    3%2 == 1当前位1，2%2当前位0，1%2当前位1，0/2当前位0

## 普通写法，便于理解

class Solution {
public:
string addBinary(string a, string b) {
int num_a = a.size() - 1;
int num_b = b.size() - 1;
string res;

int carry = 0;
while(num_a >= 0 || num_b >= 0){
int sum = 0;
sum += num_a >= 0? a[num_a--] - '0' : 0;
sum += num_b >= 0? b[num_b--] - '0' : 0;
sum += carry;
res = to_string(sum % 2) + res;
carry = sum / 2;
}
return carry == 1 ? '1' + res: res;
}
};

 

# 直接用Carry位作为求和值
class Solution {
public:
    string addBinary(string a, string b) {
        int num_a = a.size() - 1;
        int num_b = b.size() - 1;
        string res;
        int carry = 0;
        while(num_a >= 0 || num_b >= 0){
             carry += num_a >= 0? a[num_a--] - '0' : 0;
             carry += num_b >= 0? b[num_b--] - '0' : 0;
            res = to_string(carry % 2) + res;
            carry = carry / 2;
        }
        return carry == 1 ? '1' + res: res;
    }
};

LeetCode 415. Add Strings 和上面解法一样