Leetcode 刷题笔记1 单调栈part01

leetcode 739 每日温度

对于单调栈问题，我觉得是在循环外部增加一些辅助项减少时间复杂度，但增加内存空间的利用

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

leetcode 496 下一个更大元素 |

审题！！！

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        ans = [-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]]:
                    if nums2[stack[-1]] in nums1:
                        index = nums1.index(nums2[stack[-1]])
                        ans[index] = nums2[i]
                    stack.pop()
                stack.append(i)
        return ans

leetcode 503 下一个更大元素||

循环数组法一，两个相同数组串在一起：

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        ans = [-1] * len(nums)
        stack = [0]
        for i in range(len(nums) * 2):
                while len(stack) != 0 and nums[i % len(nums)] > nums[stack[-1]]:
                    ans[stack[-1]] = nums[i % len(nums)]
                    stack.pop()
                stack.append(i % len(nums))
        return ans

法二：同一个数组循环两遍