代码随想录 | 单调栈part01 part02 part03
739 每日温度
单调栈，用于快速检索某个元素左边或者右边第一个比它大或者小的元素
通过维持一个有序的栈来实现
寻找比它大的，则栈顶到栈底是递增的，反之则是递减的
class Solution:
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
res = [0]*len(temperatures)
stack = [0]
for i in range(len(temperatures)):
while len(stack) != 0 and temperatures[i] > temperatures[stack[-1]]:
res[stack[-1]] = i - stack[-1]
stack.pop()
stack.append(i)
return res
496 下一个更大元素I
使用上一题的思路，嵌入一个On的index，并使用try捕捉异常
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
stack = []
res = [-1]*len(nums1)
for i in range(len(nums2)):
while len(stack) > 0 and nums2[stack[-1]] < nums2[i]:
try:
index = nums1.index(nums2[stack[-1]])
res[index] = nums2[i]
stack.pop()
except:
stack.pop()
stack.append(i)
return res
看解答意识到还可以用dic进一步优化
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
stack = []
res = [-1]*len(nums1)
dic = {}
for i in range(len(nums1)):
dic[nums1[i]] = i
for i in range(len(nums2)):
while len(stack) > 0 and nums2[stack[-1]] < nums2[i]:
if nums2[stack[-1]] in dic:
index = dic[nums2[stack[-1]]]
res[index] = nums2[i]
stack.pop()
stack.append(i)
return res
503 下一个更大元素II
有两个思路，第一个是数组倍增两个连起来，模拟向前寻找的过程，但是需要注意储存结果时的下标
class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
res = [-1]*len(nums)
stack = []
nums2 = nums+nums
length = len(nums)
for i in range(len(nums2)):
# print(nums2[i],stack)
while len(stack) != 0 and nums2[stack[-1]] < nums2[i]:
if stack[-1] < length:
res[stack[-1]] = nums2[i]
else:
res[stack[-1]-length] = nums2[i]
stack.pop()
stack.append(i)
return res
另外一个思路是从左向右扫一遍，再从右向左扫一遍，然后结果合并
42 接雨水
确实有难度，需要仔细分析单调栈的三种情况才行
class Solution:
def trap(self, height: List[int]) -> int:
res = 0
stack = [0]
for i in range(1,len(height)):
if height[i] < height[stack[-1]]:
stack.append(i)
elif height[i] == height[stack[-1]]:
stack.pop()
stack.append(i)
else:
while len(stack)>0 and height[stack[-1]] < height[i]:
rH = height[i]
cH = height[stack[-1]]
stack.pop()
if len(stack) != 0:
lH = height[stack[-1]]
res += (min(rH,lH) - cH)*(i-stack[-1]-1)
stack.append(i)
return res
84 柱状图中最大的矩形
实际上是找左右两边第一个比它小的元素，据此确定宽度，以自己高度为高计算面积
暴力遍历？超时了。。
class Solution:
def largestRectangleArea(self, heights: List[int]) -> int:
res = -1
heights = [-1]+heights+[-1]
for i in range(1,len(heights)-1):
left = i-1
while left >= 0 and heights[left] >= heights[i]:
left -= 1
right = i+1
while right<len(heights) and heights[right] >= heights[i]:
right += 1
w = right - left - 1
res = max(res,heights[i]*w)
return res
还是得用单调栈，看了解答意识到，上一题是在找每个元素左右比它大的第一个元素，本题是在找每个元素左右比它小的第一个元素
class Solution:
def largestRectangleArea(self, heights: List[int]) -> int:
stack = [0]
res = 0
heights = [-1]+heights+[-1]
for i in range(1,len(heights)):
if heights[i] == heights[stack[-1]]:
stack.pop()
stack.append(i)
elif heights[i] > heights[stack[-1]]:
stack.append(i)
else:
while stack and heights[i] < heights[stack[-1]]:
mid = stack[-1]
stack.pop()
if stack:
left = stack[-1]
right = i
res = max(res,heights[mid]*(right-left-1))
stack.append(i)
return res
注意，这里首尾两根柱子也可以用于计算最大面积，和接雨水有区别，因此需要在首尾添加元素保证边界不会出现在首尾柱子上