LeetCode #739 DailyTempertures. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Terrible performance
Runtime: 1828 ms, faster than 5.04% of C++ online submissions for Daily Temperatures.
Memory Usage: 16.4 MB, less than 33.21% of C++ online submissions for Daily Temperatures.
LeetCode #739
Q.
Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.
주어진 매일의 온도 리스트 T 가 있는데, 각 날들이 인풋으로 들어가고, 더 따뜻한 날까지는 며칠을 기다려야 하는지를 반환한다. 해당 날 이후로 더 따뜻한 날이 없으면 0 을 반환한다.
For example, given the list of temperatures
T = [73, 74, 75, 71, 69, 72, 76, 73],
your output should be [1, 1, 4, 2, 1, 1, 0, 0].
Note:
The length of temperatures will be in the range [1, 30000].
Each temperature will be an integer in the range [30, 100].
Process
// Process
//1. Input tempertures list
//2. Iterate from begin to the end
// 2.1. Iterate from that index to the end
// 2.1.1. Check if that index temperture is cooler than future day
// 2.1.1.1. If so -> Put calculated days to resultVector
//3. Return result
// 처리과정
//1. 온도리스트 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. 해당 위치부터 끝까지 반복한다.
// 2.1.1. 해당 위치의 온도보다 따뜻한지 확인해서
// 2.1.1.1. 따뜻하면, 며칠이나 걸린 것인지 계산한 값을 리스트에 넣는다.
//3. 결과 리스트 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<int> dailyTemperatures(vector<int>& T) {
// vector<int> outputVector;
// vector<pair<int, int>> tempPair;
// for (int i = 0; i < T.size(); ++i) {
// outputVector.push_back(0);
// }
// bool isDone;
// for (int i = 0; i < T.size(); ++i) {
// for (int j = 0; j < tempPair.size(); ++j) {
// if (tempPair[j].second < T[i]) {
// outputVector.at(tempPair[j].first) = i-tempPair[j].first;
// tempPair.erase(tempPair.begin() + j);
// --j;
// }
// }
// tempPair.push_back(make_pair(i, T[i]));
// }
// return outputVector;
vector<int> outputVector;
bool isDone;
int temperture;
for (int i = 0; i < T.size(); ++i) {
isDone = false;
temperture = T[i];
for (int j = i; !isDone && j < T.size(); ++j) {
if (temperture < T[j]) {
outputVector.push_back(j-i);
isDone = true;
}
}
if (!isDone) {
outputVector.push_back(0);
}
}
return outputVector;
}
};
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