2021-10-20 Daily-Challenge
Today I have done leetcode's October LeetCoding Challenge with cpp
.
October LeetCoding Challenge 20
Description
Reverse Words in a String
Given an input string s
, reverse the order of the words.
A word is defined as a sequence of non-space characters. The words in s
will be separated by at least one space.
Return a string of the words in reverse order concatenated by a single space.
Note that s
may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.
Example 1:
Input: s = "the sky is blue"
Output: "blue is sky the"
Example 2:
Input: s = " hello world "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:
Input: s = "a good example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
Example 4:
Input: s = " Bob Loves Alice "
Output: "Alice Loves Bob"
Example 5:
Input: s = "Alice does not even like bob"
Output: "bob like even not does Alice"
Constraints:
1 <= s.length <= 104
s
contains English letters (upper-case and lower-case), digits, and spaces' '
.- There is at least one word in
s
.
Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1)
extra space?
Solution
auto speedup = [](){
cin.tie(nullptr);
cout.tie(nullptr);
ios::sync_with_stdio(false);
return 0;
}();
class Solution {
public:
string reverseWords(string s) {
reverse(s.begin(), s.end());
int len = s.length();
bool hasSpace = true;
int newLength = 0;
for(int i = 0; i < len; ++i) {
if(s[i] != ' ' || !hasSpace) {
if(s[i] == ' ') hasSpace = true;
else hasSpace = false;
s[newLength] = s[i];
newLength += 1;
}
}
if(!newLength) return "";
if(s[newLength-1] == ' ') newLength -= 1;
s.resize(newLength);
len = newLength;
int begin = -1, end = -1;
for(int i = 0; i < len; ++i) {
if(s[i] != ' ') {
if(begin == -1) begin = i;
end = i+1;
} else {
if(begin != -1 && begin != end-1) {
reverse(s.begin() + begin, s.begin() + end);
}
begin = -1;
end = -1;
}
}
if(begin != -1 && begin != end-1) reverse(s.begin() + begin, s.begin() + end);
return move(s);
}
};
// Accepted
// 57/57 cases passed (8 ms)
// Your runtime beats 42.75 % of cpp submissions
// Your memory usage beats 80.89 % of cpp submissions (7.1 MB)