【代码随想录训练营】【Day 62】【单调栈-1】| Leetcode 739, 496, 503

需强化知识点

单调栈：要寻找任一个元素的右边或者左边第一个比自己大或者小的元素的位置

题目

739. 每日温度

单调栈的基本使用：注意要判断 while len(stack) > 0

class Solution:def dailyTemperatures(self, temperatures: List[int]) -> List[int]:answer = [0] * len(temperatures)stack = []for i in range(len(temperatures)):if i == 0:stack.append(i)continueif temperatures[i] <= temperatures[stack[-1]]:stack.append(i)else:while len(stack) > 0 and temperatures[i] > temperatures[stack[-1]]:index = stack.pop()answer[index] = i - indexstack.append(i)return answer

496. 下一个更大元素 I

注意审题，nums2 是 nums1的子集，所以直接处理nums1，然后匹配即可

class Solution:def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:answer = [-1] * len(nums1)stack = [0]for i in range(1, len(nums2)):if nums2[i] <= nums2[stack[-1]]:stack.append(i)else:while len(stack) > 0 and nums2[i] > nums2[stack[-1]]:index = stack.pop()if nums2[index] in nums1:answer[nums1.index(nums2[index])] = nums2[i]stack.append(i)return answer

503. 下一个更大元素 II

循环数组的处理：直接 *2 处理

class Solution:def nextGreaterElements(self, nums: List[int]) -> List[int]:nums_new = nums + numsanswer = [-1] * len(nums) * 2stack = [0]for i in range(len(nums_new)):if nums_new[i] <= nums_new[stack[-1]]:stack.append(i)else:while len(stack) > 0 and nums_new[i] > nums_new[stack[-1]]:index = stack.pop()answer[index] = nums_new[i]stack.append(i)return answer[:len(nums)]# dp = [-1] * len(nums)# stack = []# for i in range(len(nums)*2):#     while(len(stack) != 0 and nums[i%len(nums)] > nums[stack[-1]]):#             dp[stack[-1]] = nums[i%len(nums)]#             stack.pop()#     stack.append(i%len(nums))# return dp