860. 柠檬水找零
贪心
Let's try to simulate giving change to each customer buying lemonade. Initially, we start with no five dollar bills, and no ten dollar bills.
-
If a customer brings a ＄5 bill, then we take it.
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If a customer brings a ＄10 bill, we must return a five dollar bill. If we don't have a five dollar bill, the answer is
False, since we can't make correct change. -
If a customer brings a ＄20 bill, we must return ＄15.
- If we have a ＄10 and a ＄5, then we always prefer giving change in that, because it is strictly worse for making change than three ＄5 bills.
- Otherwise, if we have three ＄5 bills, then we'll give that.
- Otherwise, we won't be able to give ＄15 in change, and the answer is
False.
class Solution:
def lemonadeChange(self, bills: List[int]) -> bool:
five, ten = 0, 0
for bill in bills:
if bill == 5:
five += 1
if bill == 10:
if five >= 1:
five -= 1
ten += 1
else:
return False
if bill == 20:
if ten >= 1 and five >= 1:
ten -= 1
five -= 1
elif five >= 3:
five -= 3
else:
return False
return True
406. 根据身高重建队列
贪心
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本题有两个维度，h和k，看到这种题目一定要想如何确定一个维度，然后在按照另一个维度重新排列。如果两个维度一起考虑一定会顾此失彼。
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按照身高h来排序呢，身高一定是从大到小排（身高相同的话则k小的站前面），让高个子在前面。
此时我们可以确定一个维度了，就是身高，前面的节点一定都比本节点高！ 那么只需要按照k为下标重新插入队列就可以了。
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局部最优：优先按身高高的people的k来插入。插入操作过后的people满足队列属性
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全局最优：最后都做完插入操作，整个队列满足题目队列属性
class Solution:
def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
# 先按照h维度的身高顺序从高到低排序。确定第一个维度
# lambda返回的是一个元组：当-x[0](维度h）相同时，再根据x[1]（维度k）从小到大排序
people.sort(key=lambda x: (-x[0], x[1]))
res = []
for p in people:
index = p[1]
res.insert(index, p)
return res
452. 用最少数量的箭引爆气球
贪心
- 局部最优：当气球出现重叠，一起射，所用弓箭最少。全局最优：把所有气球射爆所用弓箭最少。
- 为了让气球尽可能的重叠，需要对数组进行排序
# Time complexity: O(NlogN)
# Space complexity: O(N) for sorting
class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
if len(points) == 0:
return 0
points.sort(key=lambda x: x[0])
# points 不为空至少需要一支箭
res = 1
for i in range(1, len(points)):
# 气球i和气球i-1不挨着，注意这里不是>=
if points[i][0] > points[i-1][1]:
res += 1
else:
# 更新重叠气球最小右边界
points[i][1] = min(points[i][1], points[i-1][1])
return res
