代码随想录算法训练营第58天|739. 每日温度 496.下一个更大元素 I

最新推荐文章于 2025-12-02 18:04:47 发布

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文章介绍了两个使用栈数据结构解决的编程问题：一是计算每日温度中比当前日更高的温度需要等待的天数；二是找出数组中的下一个更大元素。解决方案都涉及到栈的压入和弹出操作，以及比较元素大小来更新结果。

739. 每日温度

代码随想录

class Solution(object):
    def dailyTemperatures(self, temperatures):
        """
        :type temperatures: List[int]
        :rtype: List[int]
        """
        #set up result
        result=[0]*len(temperatures)
        #set up stack, and put the first index
        stack=[0]
        # forloop from 1 to the end
        for i in range(1,len(temperatures)):
            #if current temperature is smaller than the last temperature in stack
            if temperatures[i]<=temperatures[stack[-1]]:
            #just append i to stack 
                stack.append(i)
            #if current temperature is larger than the last temperature in stack
            else:
                #while loop only when stack is not empty or 
                #current temperature is larger than the last temperature in stack
                while len(stack)!=0 and temperatures[i]>temperatures[stack[-1]]:
                    #record result
                    result[stack[-1]]=i-stack[-1]
                    #pop last element in stack 
                    stack.pop()
                stack.append(i)
        return result

496.下一个更大元素 I

代码随想录

class Solution(object):
    def nextGreaterElement(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        #set up result, a list of -1, which means no match
        result=[-1]*len(nums1)
        #set up stack
        stack=[0]
        #loop from 1 
        for i in range(1,len(nums2)):
            #if current nums value is smaller or equal to the smallest value 
            #recorded in stack
            if nums2[i]<=nums2[stack[-1]]:
                #append i to stack
                stack.append(i)
            #if current nums value is greater the smallest value 
            #recorded in stack
            else:
                #loop when stack is not empty and the current nums value 
                #is greater the smallest value recorded in stack
                while len(stack)!=0 and nums2[i]>nums2[stack[-1]]:
                    #if current smallest value recorded in stack is in nums1
                    if nums2[stack[-1]] in nums1:
                        #update result
                        result[nums1.index(nums2[stack[-1]])]=nums2[i]
                    #pop usded stack element
                    stack.pop()
                #append new i
                stack.append(i)
        return result