344. Reverse String
輸入一個字符串，輸出反向
Example:
Given s = "hello", return "olleh".
我的代碼：怎麼還能有這種題
class Solution(object):
def reverseString(self, s):
"""
:type s: str
:rtype: str
"""
return s[::-1]496. Next Greater Element I
給出一個數組，並給出一個子集，對子集中每個數字，找到原數組中的位置，向右找第一個比這個數字大的數，沒有則輸出-1
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1.我的代碼：
class Solution(object):
def nextGreaterElement(self, findNums, nums):
"""
:type findNums: List[int]
:type nums: List[int]
:rtype: List[int]
"""
ans = []
for i in findNums:
index = nums.index(i)
for j in nums[index:]:
if j > i:
ans.append(j)
break
elif j == nums[-1]:
ans.append(-1)
return ans
d = {}
st = []
ans = []
for x in nums:
while len(st) and st[-1] < x:
d[st.pop()] = x
st.append(x)
for x in findNums:
ans.append(d.get(x, -1))
return ans計算河岸的邊長
Example:
[[0,1,0,0], [1,1,1,0], [0,1,0,0], [1,1,0,0]] Answer: 16 Explanation: The perimeter is the 16 yellow stripes in the image below:
我的代碼：
class Solution(object):
def islandPerimeter(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
count = 0
for i in range(len(grid)):
for j in range(len(grid[i])):
if grid[i][j] == 1:
count += 4
if i != 0 and grid[i-1][j] == 1:
count -= 1
if i != len(grid) - 1 and grid[i+1][j] == 1:
count -= 1
if j != 0 and grid[i][j-1] == 1:
count -= 1
if j != len(grid[i])-1 and grid[i][j+1] == 1:
count -= 1
return countdef islandPerimeter(self, grid):
return sum(sum(map(operator.ne, [0] + row, row + [0]))for row in grid + map(list, zip(*grid)))實現matlab的reshape函數
Example 1:
Input: nums = [[1,2], [3,4]] r = 1, c = 4 Output: [[1,2,3,4]] Explanation:
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
Example 2:
Input: nums = [[1,2], [3,4]] r = 2, c = 4 Output: [[1,2], [3,4]] Explanation:我的代碼
There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
class Solution(object):
def matrixReshape(self, nums, r, c):
"""
:type nums: List[List[int]]
:type r: int
:type c: int
:rtype: List[List[int]]
"""
lists = []
for m in nums:
for n in m:
lists.append(n)
if len(lists) != r*c:
return nums
ans = []
for i in range(r):
ans.append([])
for j in range(c):
ans[i].append(lists.pop(0))
return ansdef matrixReshape(self, nums, r, c):
return nums if len(sum(nums, [])) != r * c else map(list, zip(*([iter(sum(nums, []))]*c)))
def matrixReshape(self, nums, r, c):
flat = sum(nums, [])
if len(flat) != r * c:
return nums
tuples = zip(*([iter(flat)] * c))
return map(list, tuples)
def matrixReshape(self, nums, r, c):
if r * c != len(nums) * len(nums[0]):
return nums
it = itertools.chain(*nums)
return [list(itertools.islice(it, c)) for _ in xrange(r)]412. Fizz Buzz
輸入一個正整數，輸出一個列表爲1~這個數，其中被3整除的，被5整除的，以及又被3又被5整除的用其他文字替換
Example:
n = 15,
Return:
[
"1",
"2",
"Fizz",
"4",
"Buzz",
"Fizz",
"7",
"8",
"Fizz",
"Buzz",
"11",
"Fizz",
"13",
"14",
"FizzBuzz"
]
我的代碼：沒什麼難度，想到什麼就寫什麼了class Solution(object):
def fizzBuzz(self, n):
"""
:type n: int
:rtype: List[str]
"""
ans = []
for i in range (n):
if (i+1) % 3 == 0 and (i+1) % 5 == 0:
ans.append('FizzBuzz')
elif (i+1) % 3 == 0:
ans.append('Fizz')
elif (i+1) % 5 == 0:
ans.append('Buzz')
else:
ans.append(str(i+1))
return ans
def fizzBuzz(self, n):
return ['Fizz' * (not i % 3) + 'Buzz' * (not i % 5) or str(i) for i in range(1, n+1)]682. Baseball Game
給出一個序列，按要求積分，數字則直接記有效分數，+記上兩個有效分數的和爲此次有效分數，D將上個有效分數乘2爲此次有效分數，C清除上一個有效分數
最後給出總有效分
Example 1:
Input: ["5","2","C","D","+"] Output: 30 Explanation: Round 1: You could get 5 points. The sum is: 5. Round 2: You could get 2 points. The sum is: 7. Operation 1: The round 2's data was invalid. The sum is: 5. Round 3: You could get 10 points (the round 2's data has been removed). The sum is: 15. Round 4: You could get 5 + 10 = 15 points. The sum is: 30.
Example 2:
Input: ["5","-2","4","C","D","9","+","+"] Output: 27 Explanation: Round 1: You could get 5 points. The sum is: 5. Round 2: You could get -2 points. The sum is: 3. Round 3: You could get 4 points. The sum is: 7. Operation 1: The round 3's data is invalid. The sum is: 3. Round 4: You could get -4 points (the round 3's data has been removed). The sum is: -1. Round 5: You could get 9 points. The sum is: 8. Round 6: You could get -4 + 9 = 5 points. The sum is 13. Round 7: You could get 9 + 5 = 14 points. The sum is 27.我的代碼：也是想到啥寫啥了
class Solution(object):
def calPoints(self, ops):
"""
:type ops: List[str]
:rtype: int
"""
anslist = []
for i in ops:
if i == 'C':
anslist.pop()
elif i == 'D':
anslist.append(anslist[-1]*2)
elif i == '+':
anslist.append(anslist[-2]+anslist[-1])
elif isinstance(int(i),int):
anslist.append(int(i))
return sum(anslist)