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61 lines
1.7 KiB
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61 lines
1.7 KiB
Plaintext
---
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title: '739. Daily Temperatures'
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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 i^th day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead
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icon: dot
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sidebar:
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label: 'Daily Temperatures'
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badge: 'Medium'
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---
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<Badge variant="accent">Array</Badge>
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### Example 1:
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- Input: `temperatures = [73,74,75,71,69,72,76,73]`
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- Output: `[1,1,4,2,1,1,0,0]`
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### Example 2:
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- Input: `temperatures = [30,40,50,60]`
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- Output: `[1,1,1,0]`
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### Example 3:
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- Input: `temperatures = [30,60,90]`
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- Output: `[1,1,0]`
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### Constraints:
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- `1 <= temperatures.length <= 10^5`
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- `30 <= temperatures[i] <= 100`
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## Approach
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```mermaid
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flowchart TD
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S(["dailyTemperatures(temperatures)"]) --> I["ans = zeros(n), stack = [] — indices of days still waiting"]
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I --> L{"more (i, temp)?"}
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L -- no --> E(["return ans"])
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L -- yes --> W{"stack non-empty and temperatures[stack[-1]] < temp?"}
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W -- yes --> P["prev = stack.pop(); ans[prev] = i - prev — today resolves that day"]
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P --> W
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W -- no --> A["stack.append(i)"]
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A --> L
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```
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## Solution
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```py
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class Solution:
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def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
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ans = [0] * len(temperatures)
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stack = []
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for i, temp in enumerate(temperatures):
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while stack and temperatures[stack[-1]] < temp:
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prev = stack.pop()
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ans[prev] = i - prev
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stack.append(i)
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return ans
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```
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## Explanation
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<YouTube id="cTBiBSnjO3c" title="Daily Temperatures - Monotonic Stack - Leetcode 739 - Python" />
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