- Leetcode 43. Multiply Strings
- Leetcode 49. Group Anagrams
- Leetcode 67. Add Binary
- Leetcode 539. Minimum Time Difference
Leetcode String Series II
Leetcode 43. Multiply Strings
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
Note:
-
The length of both num1 and num2 is < 110.
-
Both num1 and num2 contains only digits 0-9.
-
Both num1 and num2 does not contain any leading zero.
-
You must not use any built-in BigInteger library or convert the inputs to integer directly.
class Solution {
public:
string multiply(string num1, string num2) {
const int m = num1.size();
const int n = num2.size();
string res(m+n+1, '0');
for (int i=m-1; i>=0; i--)
for (int j=n-1; j>=0; j--) {
int temp = (num1[i]-'0')*(num2[j]-'0') + (res[i+j+2]-'0');
res[i+j+2] = '0'+temp%10;
res[i+j+1] += temp/10;
}
size_t begin = res.find_first_not_of('0');
if(begin == string::npos) return "0";
return res.substr(begin);
}
};
Leetcode 49. Group Anagrams
Given an array of strings, group anagrams together.
For example, given: [“eat”, “tea”, “tan”, “ate”, “nat”, “bat”],
Return:
[
["ate", "eat","tea"],
["nat","tan"],
["bat"]
]
Note: All inputs will be in lower-case.
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
vector<vector<string>> res;
if (strs.empty()) return res;
unordered_map<string, vector<string>> table;
for (string str: strs) {
string key = str;
sort(key.begin(), key.end());
table[key].push_back(str);
}
for (auto i=table.begin(); i!=table.end(); i++)
res.push_back(i->second);
return res;
}
};
Leetcode 67. Add Binary
Given two binary strings, return their sum (also a binary string).
For example,
a = “11”
b = “1”
Return “100”.
class Solution {
public:
string addBinary(string a, string b) {
if (a.empty()) return b;
if (b.empty()) return a;
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int i = 0, j = 0, carry = 0;
string res = "";
while (i<a.size() || j<b.size()) {
if (i<a.size())
carry += a[i++]-'0';
if (j<b.size())
carry += b[j++]-'0';
res += (char)(carry%2+'0');
carry /= 2;
}
if (carry) res += '1';
reverse(res.begin(), res.end());
return res;
}
};
Leetcode 539. Minimum Time Difference
Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.
Example 1:
Input: ["23:59","00:00"]
Output: 1
Note:
-
The number of time points in the given list is at least 2 and won’t exceed 20000.
-
The input time is legal and ranges from 00:00 to 23:59.
class Solution {
public:
int findMinDifference(vector<string>& timePoints) {
if (timePoints.empty()) return 0;
sort(timePoints.begin(), timePoints.end());
int res = INT_MAX;
const int n = timePoints.size();
int val[n] = {0};
int val2[n] = {0};
for (int i = 0; i<timePoints.size(); i++) {
val[i] = stoi(timePoints[i].substr(0,2));
val2[i] = stoi(timePoints[i].substr(3,2));
}
for (int i = 1; i<timePoints.size(); i++) {
res = min(res, (val[i]-val[i-1])*60+val2[i]-val2[i-1]);
res = min(res, (val[i-1]+24-val[i])*60+val2[i-1]-val2[i]);
}
res = min(res, (val[0]+24-val[n-1])*60+val2[0]-val2[n-1]);
return res;
}
};
近期评论