739 Daily Temperatures 每日温度
Description:
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:
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 <= 10^5
30 <= temperatures[i] <= 100
题目描述:
请根据每日 气温 列表，重新生成一个列表。对应位置的输出为：要想观测到更高的气温，至少需要等待的天数。如果气温在这之后都不会升高，请在该位置用 0 来代替。
示例 :
例如，给定一个列表 temperatures = [73, 74, 75, 71, 69, 72, 76, 73]，你的输出应该是 [1, 1, 4, 2, 1, 1, 0, 0]。
提示:
气温 列表长度的范围是 [1, 30000]。每个气温的值的均为华氏度，都是在 [30, 100] 范围内的整数。
思路:
单调栈
遍历的时候维护一个单调递减的元素下标的单调栈
每次取到较大元素就弹出栈中元素并记录距离
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
class Solution
{
public:
vector<int> dailyTemperatures(vector<int>& temperatures)
{
stack<int> s;
int n = temperatures.size();
vector<int> result(n);
for (int i = 0; i < n; i++)
{
while (!s.empty() and temperatures[s.top()] < temperatures[i])
{
result[s.top()] = i - s.top();
s.pop();
}
s.push(i);
}
return result;
}
};
Java:
class Solution {
public int[] dailyTemperatures(int[] temperatures) {
Stack<Integer> stack = new Stack<>();
int n = temperatures.length;
int result[] = new int[n];
for (int i = 0; i < n; i++) {
while (!stack.isEmpty() && temperatures[stack.peek()] < temperatures[i]) result[stack.peek()] = i - stack.pop();
stack.push(i);
}
return result;
}
}
Python:
class Solution:
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
stack, result = [], [0] * (n := len(temperatures))
for i, t in enumerate(temperatures):
while stack and temperatures[stack[-1]] < t:
result[stack.pop()] = i - stack[-1]
stack.append(i)
return result
