LeetCode初级算法的Python实现--数组
LeetCode初级算法的Python实现--数组
# -*- coding: utf-8 -*-
"""
@Created on 2018/6/3 17:06
@author: ZhifengFang
"""
# 排列数组删除重复项
def removeDuplicates(nums):
if len(nums) <= 1:
return len(nums)
i = 1
while len(nums) != i:
if nums[i] == nums[i - 1]:
del nums[i]
i -= 1
i += 1
return len(nums)
# 买卖股票最佳时机2
def maxProfit(prices):
max = 0
if len(prices) <= 1:
return 0
for i in range(len(prices) - 1):
if prices[i] < prices[i + 1]:
max += prices[i + 1] - prices[i]
return max
# 旋转数组
def rotate(nums, k):
# nums = nums[-k:] + nums[:k + 1]
# print(nums)
if len(nums) > 1:
k = k % len(nums)
if k != 0:
temp = nums[-k:]
nums[k:] = nums[:len(nums) - k]
nums[0:k] = temp
print(nums)
# 判断数组中是否有重复元素
def containsDuplicate(nums):
# if len(nums)>len(set(nums)):
# return True
# return False
for num in nums:
if nums.count(num) > 1:
return True
return False
# 获得里面只出现一次的数字
def singleNumber(nums):
numCounts = {}
result = []
for num in nums:
numCounts[num] = numCounts.get(num, 0) + 1
for key in numCounts.keys():
if numCounts.get(key) == 1:
result.append(key)
break
return result[0]
# 两个数组的交集 II
def intersect(nums1, nums2):
if len(nums2) < len(nums1):
nums1, nums2 = nums2, nums1
newNums = []
i = 0
while i < len(nums1):
j = 0
while j < len(nums2):
if nums1[i] == nums2[j]:
newNums.append(nums2[j])
del nums1[i], nums2[j]
i -= 1
j -= 1
break
j += 1
i += 1
return newNums
# print(intersect([9],[7,8,3,9,0,0,9,1,5]))
# 加1
def plusOne(digits):
strDigits = ''
for example in digits:
strDigits += str(example)
strDigits = int(strDigits) + 1
listDigits = [int(str) for str in str(strDigits)]
return listDigits
# print(plusOne([1, 2, 3]))
# 移动0
def moveZeroes(nums):
# for i in range(len(nums)):
i = 0
zeroesCount = 0
while i + zeroesCount < len(nums):
if nums[i] == 0:
nums[i:] = nums[i + 1:] + [0]
i -= 1
zeroesCount += 1
i += 1
return nums
# 两数和
def twoSum(nums, target):
d = {}
for x in range(len(nums)):
a = target - nums[x]
if nums[x] in d:
return d[nums[x]], x
else:
d[a] = x
nums = [3, 2, 4]
target = 6
# print(twoSum(nums, target))
def isXT(strs):
strSet = set(strs)
for s in strSet:
if s != ".":
if strs.count(s) > 1:
return False
return True
# 有效的数独
def isValidSudoku(board):
for i in range(9):
boardLie = [example[i] for example in board]
key1 = int(i / 3) * 3 + 1
key2 = 1 + (i % 3) * 3
boardGe = [board[key1 - 1][key2 - 1], board[key1 - 1][key2], board[key1 - 1][key2 + 1],
board[key1][key2 - 1], board[key1][key2], board[key1][key2 + 1],
board[key1 + 1][key2 - 1], board[key1 + 1][key2], board[key1 + 1][key2 + 1]]
if isXT(board[i]) == False:
return False
if isXT(boardLie) == False:
return False
if isXT(boardGe) == False:
return False
return True
board = [[".", ".", "4", ".", ".", ".", "6", "3", "."],
[".", ".", ".", ".", ".", ".", ".", ".", "."],
["5", ".", ".", ".", ".", ".", ".", "9", "."],
[".", ".", ".", "5", "6", ".", ".", ".", "."],
["4", ".", "3", ".", ".", ".", ".", ".", "1"],
[".", ".", ".", "7", ".", ".", ".", ".", "."],
[".", ".", ".", "5", ".", ".", ".", ".", "."],
[".", ".", ".", ".", ".", ".", ".", ".", "."],
[".", ".", ".", ".", ".", ".", ".", ".", "."]]
# print(isValidSudoku(board))
# 旋转图像
def rotate(matrix):
for i in range(len(matrix)):
for j in range(i+1,len(matrix)):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
matrix[i].reverse()
print(matrix)
ma = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
rotate(ma)
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