1. 移動零
class Solution(object):
def moveZeroes(self, nums):
"""
:type nums: List[int]
:rtype: None Do not return anything, modify nums in-place instead.
聰明解法
1. 兩個指針，i指針迭代，j指針表示（從左往右）非零元素個數的索引，i遇到非零就複製給j
即：把非0元素都往數組前面扔，數組後面都應該是0
https://leetcode-cn.com/problems/move-zeroes/solution/dong-hua-yan-shi-283yi-dong-ling-by-wang_ni_ma/
2. i不是扔，是交換
"""
# j = 0
# for i in xrange(len(nums)):
# if nums[i] != 0:
# nums[j] = nums[i]
# j += 1
# for k in xrange(j, len(nums)):
# nums[k] = 0
j = 0
for i in range(len(nums)):
if nums[i] != 0:
nums[i], nums[j] = nums[j], nums[i]
j += 1
2. 環形鏈表
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
笨辦法
有環說明一個節點被指向了兩次
使用set存儲節點索引
聰明法
節點被訪問過，就把值改爲True
雙指針，一個快指針、一個慢指針，總會相遇
"""
if not head:
return False
slow, fast = head, head
while slow and fast:
slow = slow.next
if fast.next:
fast = fast.next.next
else:
return False
if slow == fast:
return True
return False
3. 迴文鏈表
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
笨辦法
鏈表輸出到數組，然後兩個指針前後移動檢測數組元素是否相等
時間O(2N)，空間O(N)
"""
if not head:
False
array = []
node = head
while node:
array.append(node.val)
node = node.next
i, j = 0, len(array) - 1
while i <= j:
if array[i] != array[j]:
return False
i += 1
j -= 1
return True
4. 尋找重複數
287. find-the-duplicate-number
class Solution(object):
def findDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
笨辦法
暴力搜索
哈希表, 統計元素出現次數, 大於1就返回
聰明辦法
快慢指針法
可以將數組表示成一個帶環的圖: https://pic.leetcode-cn.com/f9104ccbbd4e572a8a9d6b2e7e2f75084b91c045b48e1dfc65cee238460257a1-image.png
快慢指針都從起始位置出發, 慢指針一次走1步, 快指針一次走2步, 直到相遇
然後, 慢指針回到起始位置, 快指針在相遇位置, 兩者同時一次走1步, 再次相遇的點即我們要的數
"""
slow, fast = 0, 0
while slow == fast == 0 or slow != fast:
slow, fast = nums[slow], nums[nums[fast]]
slow = 0
while slow != fast:
slow, fast = nums[slow], nums[fast]
return slow
5. 盛最多水的容器
class Solution(object):
def maxArea(self, height):
"""
:type height: List[int]
:rtype: int
笨辦法
暴力搜索
聰明辦法
雙指針法
左右指針分別表示容器的左右邊界, 初始值分別爲樣本的左右邊界
當左指針小於右指針時, 左指針右移; 否則, 右指針左移
每次移動時, 計算當前面積, 直到左右指針相遇, 即可得最大面積
https://leetcode-cn.com/problems/container-with-most-water/solution/sheng-zui-duo-shui-de-rong-qi-by-leetcode-solution/
"""
max_area = 0
left, right = 0, len(height) - 1
while left < right:
area = (right - left) * min(height[left], height[right])
max_area = max(max_area, area)
if height[left] < height[right]:
left += 1
else:
right -= 1
return max_area
6. 相交鏈表
160. intersection-of-two-linked-lists
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
笨辦法
固定A的某個節點，遍歷B的所有節點，相等返回；
遍歷A的所有節點執行上面操作；
聰明辦法
A、B同時開始往後走；
A走完了走B；B走完了走A；
若相遇則返回，否則不相交；
"""
pA, pB = headA, headB
is_pa_linked, is_pb_linked = False, False
while pA and pB:
if pA == pB:
return pA
pA, pB= pA.next, pB.next
if not pA and not is_pa_linked:
pA = headB
is_pa_linked = True
if not pB and not is_pb_linked:
pB = headA
is_pb_linked = True
return None
7. 迴文鏈表
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
思路
鏈表輸出到數組，然後兩個指針前後移動檢測數組元素是否相等
時間O(2N)，空間O(N)
"""
if not head:
False
array = []
node = head
while node:
array.append(node.val)
node = node.next
i, j = 0, len(array) - 1
while i <= j:
if array[i] != array[j]:
return False
i += 1
j -= 1
return True