Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.
Example 1:
Input: temperatures = [73,74,75,71,69,72,76,73]
Output: [1,1,4,2,1,1,0,0]
Example 2:
Input: temperatures = [30,40,50,60]
Output: [1,1,1,0]
Example 3:
Input: temperatures = [30,60,90]
Output: [1,1,0]
Constraints:
1 <= temperatures.length <= 105
30 <= temperatures[i] <= 100
class Solution {
public:
vector<int> dailyTemperatures(vector<int>& temperatures) {
int n = temperatures.size();
stack<int> st;
vector<int> res(n,0);
st.push(0);
for(int i = 1;i < n;i++){
if(temperatures[i] <= temperatures[st.top()]) st.push(i);
else{
while(!st.empty() && temperatures[i] > temperatures[st.top()]){
int j = st.top();
st.pop();
res[j] = i - j;
}
st.push(i);
}
}
return res;
}
};
for循环中的if语句还可以进一步精简,直接将if-else去除