leetcode-python 第四周

1.The Best To Buy and Sell Stock II [60ms]

#方法1：后来经过测试，题目是可以当天卖出然后当天再买入的
#这样问题就可以化解为贪心
#例如[1, 2, 5, 6], 画个数轴就知道了
class Solution(object):
def maxProfit(self, prices):
ans = 0
for i in range(len(prices)-1):
if prices[i+1] - prices[i] > 0:
ans += prices[i+1] - prices[i]
return ans

2.Find Minimum in Rotated Sorted Array II[48ms]

#方法1：二分搜索，如果其中一边是正序的话，取最左边的数
#再去另外一边看有没有比这个数更小的
class Solution(object):
def findMin(self, nums):
ans = 100000
low = 0
high = len(nums)
while low != high :
mid = int((low+high) / 2)
if nums[low] < nums[mid] :
ans = min(nums[low], ans)
low = mid + 1
else:
ans = min(nums[mid], ans)
high = mid
return ans

3.Find Peek Element [52ms]

#方法1：注意题目是找到任意一个就可以而且左右元素不相等。
#比较n次，最大的元素一定是波峰 [52ms]
#方法2：二分查找，如果中间是上升则往右边找。[56ms]
class Solution(object):
def findPeakElement(self, nums):
ans = 0
for i in range(1, len(nums)):
if nums[i] >= nums[i-1]:
ans = i
return ans

class Solution(object):
def findPeakElement(self, nums):
low, high = 0, len(nums)-1
while low <= high :
if low == high :
return low
mid = (int(low+high) / 2)
if nums[mid] < nums[mid+1] :
low = mid + 1
else :
high = mid

4.Minimum Size Subarray Sum [65ms]

#方法1：自己写的滑窗，因为不用一头一尾的while循环，
#所以需要额外判断的条件比较多
#方法2：二分枚举，[96ms]
class Solution(object):
def minSubArrayLen(self, s, nums) :
low = 0
addup = 0
ans = 100000

for i in range(len(nums)):
addup += nums[i]

if addup >= s :
ans = min(ans, i-low+1)

while addup >= s and addup - nums[low] >= s :
addup -= nums[low]
low += 1
ans = min(ans, i-low+1)
if addup >= s:
ans = min(ans, len(nums)-low)

if ans == 100000 :
return 0
return ans

class Solution:
def minSubArrayLen(self, s, nums):
size = len(nums)
left, right = 0, size
bestAns = size + 1
while left <= right:
mid = int((left + right) / 2)
if self.solve(mid, s, nums):
bestAns = mid #bestAns也代表窗口大小，随二分不断试试减少
right = mid - 1
else:
left = mid + 1
return [0, bestAns][bestAns <= size]
def solve(self, l, s, nums):
sums = 0
for x in range(len(nums)):
sums += nums[x]
if x >= l:
sums -= nums[x - l] #窗口滑动
if sums >= s:
return True
return False

5.Subset [108ms]

class Solution(object):
import copy
def subsets(self, nums):
result = []
tmp = []
self.dfs(0, tmp, result, nums)
return result

def dfs(self, k, tmp, result, nums) :
r = copy.deepcopy(tmp)
result.append(r)
for i in range(k, len(nums)) :
tmp.append(nums[i])
self.dfs(i+1, tmp, result, nums)
del tmp[-1]
return

6.Subset II [124ms]

#方法1:例子[1, 2, 2]
class Solution(object):
import copy

def dfs(self, k, tmp, result, nums) :
r = copy.deepcopy(tmp)
result.append(r)
for i in range(k, len(nums)) :
#防止重复
if i > k and nums[i] == nums[i-1] :
continue
tmp.append(nums[i])
self.dfs(i+1, tmp, result, nums)
del tmp[-1]
return

def subsetsWithDup(self, nums) :
result = []
tmp = []
nums.sort()
self.dfs(0, tmp, result, nums)
return result

7.Triangle [54ms]

#方法1：动态规划的思想，从下往上，改变三角形的数
#取下面两个接邻数的最小
# f[i][j] = min(f[i+1][j], f[i+1][j+1]) + f[i][j]
class Solution(object):
def minimumTotal(self, triangle):
for i in range(len(triangle)-2, -1, -1) :
for j in range(i+1) :
triangle[i][j] = min(triangle[i+1][j], triangle[i+1][j+1]) + triangle[i][j]
return triangle[0][0]