def prevGreaterGap(self, arr: list):
"""
左边第一个最大元素的距离
[2, 1, 2, 4, 3]返回结果[0, 1, 0, 0, 1]
"""
result: list = [None] * len(arr)
stack = Stack()
for i in range(len(arr)):
while (not stack.isEmpty() and arr[stack.peek()] <= arr[i]):
stack.pop()
result[i] = i - stack.peek() if not stack.isEmpty() else 0
stack.push(i)
return result
def anticlockwiseNextGreaterElement(self, arr: list):
"""
循环数组下的, 左边最大元素, 即逆时针方向的第一个最大元素
[2, 1, 2, 4, 3]返回结果[3, 2, 3, -1, 4]
"""
result: list = [None] * len(arr)
stack = Stack()
n = len(arr)
for i in range(len(arr) * 2):
while (not stack.isEmpty() and stack.peek() <= arr[i % n]):
stack.pop()
result[i % n] = stack.peek() if not stack.isEmpty() else -1
stack.push(arr[i % n])
return result
def prevGreaterElement(self, arr: list):
"""
左边第一个比元素大的值
[2, 1, 2, 4, 3]返回结果[-1, 2, -1, -1, 4]
"""
result: list = [None] * len(arr)
stack = Stack()
for i in range(len(arr)):
while (not stack.isEmpty() and stack.peek() <= arr[i]):
stack.pop()
result[i] = stack.peek() if not stack.isEmpty() else -1
stack.push(arr[i])
return result
def clockwiseNextGreaterElement(self, arr: list):
"""
循环数组下的, 右边最大元素, 即顺时针方向的第一个最大元素
[2, 1, 2, 4, 3]返回结果[4, 2, 4, -1, 4]
"""
result: list = [None] * len(arr)
stack = Stack()
n: int = len(arr)
# 假设数组长度*2, 使用%, 其实还是一倍长
for i in range(2 * len(arr) - 1, -1, -1):
while (not stack.isEmpty() and stack.peek() <= arr[i % n]):
stack.pop()
result[i % n] = stack.peek() if not stack.isEmpty() else -1
stack.push(arr[i % n])
return result
class StackToQueue:
def __init__(self):
self.stack1 = Stack()
self.stack2 = Stack()
def push(self, val):
"""push进stack1"""
self.stack1.push(val)
def pop(self):
self.peek()
return self.stack2.pop()
def peek(self):
"""stack1的元素push进stack2"""
if self.stack2.isEmpty():
while not self.stack1.isEmpty():
self.stack2.push(self.stack1.pop())
return self.stack2.peek()
def isEmpty(self):
return False if (not self.stack1.isEmpty()
or not self.stack2.isEmpty()) else True
def printQueue(self):
while not self.isEmpty():
print(self.pop(), end="\t")
def nextGreaterGap(self, arr: list):
"""
[2, 1, 2, 4, 3]返回结果[3, 1, 1, 0, 0]
从左到右:
比2大的为右边第3个
比1大的为右边第1个
比2大的为右边第1个
比4大的为右边第0个, 即没有
比3大的为右边第0个, 即没有
"""
result: list = [None] * len(arr)
stack = Stack()
for i in range(len(arr) - 1, -1, -1):
while (not stack.isEmpty()
and arr[stack.peek()] <= arr[i]): # 操作索引
stack.pop()
# 这里stack[-1]为右边第一个最大值的索引
result[i] = stack.peek() - i if not stack.isEmpty() else 0
stack.push(i) # 索引入栈
return result
def nextGreaterElement(self, arr: list):
"""
右边第一个比元素大的值
[2, 1, 2, 4, 3]返回结果[4, 2, 4, -1, -1]
从左到右:
比2大的第一个为4
比1大的第一个为2
比2大的第一个为4
比4大的第一个没有-1
比3大的第一个没有-1
时间复杂度: O(n)
"""
result: list = [None] * len(arr)
stack = Stack()
for i in range(len(arr) - 1, -1, -1):
while (not stack.isEmpty() and stack.peek() <= arr[i]):
stack.pop()
result[i] = stack.peek(
) if not stack.isEmpty() else -1 # 这里stack[-1]为右边第一个最大值
stack.push(arr[i]) # 值入栈
return result