134. 加油站
思路：
画图，sum+=gas[i]-cost[i]，然后找到最低点的下一个位置就是
total代表途中的油量，最低点后油量开始变多了，
class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
total = 0
min_val = 0
start = 0
for i in range(len(gas)):
total += (gas[i] - cost[i])
if total < min_val:
min_val = total
start = i + 1
if total<0:
return -1
return 0 if start == len(gas) else start
135. 分发糖果
思路：
1 从左往右，如果r[i]>r[i-1] left[i]=left[i-1] + 1 left[0]=1
2 从右往左，如果r[i]>r[i+1] right[i] = right[i+1] + 1 right[n-1]=1
class Solution:
def candy(self, ratings: List[int]) -> int:
if not ratings: return 0
n = len(ratings)
left = [1] * len(ratings)
# right = [1] * len(ratings)
right = 1
result = 0 # 最小糖果数量
# 左规则
for i in range(1, len(ratings)):
if ratings[i] > ratings[i-1]:
left[i] = left[i-1] + 1
# 右规则
for i in range(n-1,-1,-1):
if i<n-1 and ratings[i] > ratings[i+1]:
# right[i] = right[i+1] + 1
right += 1
else:
right = 1
# result += max(left[i], right[i])
result += max(left[i], right)
return result
860. 柠檬水找零
思路：
情况一：账单是5，直接收下。
情况二：账单是10，消耗一个5，增加一个10
情况三：账单是20，优先消耗一个10和一个5，如果不够，再消耗三个5
class Solution:
def lemonadeChange(self, bills: List[int]) -> bool:
five, ten, twenty = 0, 0, 0
for bill in bills:
if bill == 5:
five += 1
elif bill == 10:
# 需要一个5
if five > 0:
five -= 1
ten += 1
else:
return False
elif bill == 20:
# 两种情况：1 取1个10和一个5（10只能给20找零，优先使用）； 2 取3个5
if ten>0 and five > 0:
ten -= 1
five -= 1
twenty += 1
elif five>=3:
five -= 3
twenty += 1
else:
return False
return True
406. 根据身高重建队列
思路：
1 先按照升高h降序，个数k升序排序
2 迭代每个人，按照k作为索引插入
class Solution:
def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
people.sort(key = lambda x: (-x[0], x[1]))
queue = []
for p in people:
queue.insert(p[1], p)
return queue
