反转字符串
思路
无难度，双指针即可，注意特殊条件和循环终止条件
代码
class Solution(object):
def reverseString(self, s):
"""
:type s: List[str]
:rtype: None Do not return anything, modify s in-place instead.
"""
n = len(s)
if n > 1:
head, tail = 0, n - 1
while head < tail:
s[head], s[tail] = s[tail], s[head]
head += 1
tail -= 1
柠檬水找零
思路
本题思路比较直接，不用考虑太多迭代递归等等，直接遍历即可，需要注意：
- 找零优先使用较大币值
代码
class Solution(object):
def lemonadeChange(self, bills):
"""
:type bills: List[int]
:rtype: bool
"""
remain_cash = dict()
remain_cash[5] = 0
remain_cash[10] = 0
remain_cash[20] = 0
for bill in bills:
if bill == 5:
remain_cash[5] += 1
elif bill == 10:
if remain_cash[5] <= 0:
return False
else:
remain_cash[5] -= 1
remain_cash[10] += 1
else:
if remain_cash[10] > 0 and remain_cash[5] > 0:
remain_cash[10] -= 1
remain_cash[5] -= 1
elif remain_cash[5] >= 3:
remain_cash[5] -= 3
else:
return False
remain_cash[20] += 1
return True
买卖股票的最佳时机 II
思路
当天可售出再买回，因此用贪心算法即可
代码
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
result = 0
for i in range(1, len(prices)):
profit = prices[i] - prices[i-1]
if profit > 0:
result += profit
return result
