1. 合并列表中字典字段

如下两个列表，需要将oldList转化为newList，去掉相同字段的字典，并且去掉的参数里面的值要相加。

oldList = [{'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1972}, {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0}, {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1450}, {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0}, {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1334}] newList = [{'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 4756}, {'3-3': 406, '3-2': 0, '3-1': 0, '3-0': 0}]

以下代码 A、B、C、D四选一：

==【A】================================================================================
import operator
oldList = [{'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1972},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0//0': 0, '0//1': 0, '0//2': 0, '0//3': 1450},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1334}]
newList = []
newList.append(oldList[0])
for t in range(1,len(oldList)):
    for li in newList:
        if operator.eq(li.keys(), oldList[t].keys()):
            for key in li.keys():
                li[key] += oldList[t][key]
            break
        elif operator.eq(li,newList[-1]):
            newList.append(oldList[t])
            break
print(newList)

==【B】================================================================================
import operator
oldList = [{'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1972},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1450},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1334}]
newList = []
newList.append(oldList[0])
for t in range(1,len(oldList)):
    for li in newList:
        if operator.eq(li.keys(), oldList[t].keys()):
            for key in li.keys():
                li[key] += oldList[t][key]
            break
        elif operator.eq(li,newList[-1]):
            newList.append(oldList[t])
            break
print(newList)

==【C】================================================================================
import operator
oldList = [{'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1972},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0*0': 0, '0*1': 0, '0*2': 0, '0*3': 1450},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1334}]
newList = []
newList.append(oldList[0])
for t in range(1,len(oldList)):
    for li in newList:
        if operator.eq(li.keys(), oldList[t].keys()):
            for key in li.keys():
                li[key] += oldList[t][key]
            break
        elif operator.eq(li,newList[-1]):
            newList.append(oldList[t])
            break
print(newList)

==【D】================================================================================
import operator
oldList = [{'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1972},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1450},
           {'3-3': 203, '3-2': 0, '3-1': 0, '3-0': 0},
           {'0-0': 0, '0-1': 0, '0-2': 0, '0-3': 1334}]
newList = []
newList.append(oldList[0])
for t in range(1,len(oldList)):
    for li in newList:
        if operator.eq(li.keys(), oldList[t].keys()):
            for key in li.keys():
                li[key] += oldList[t][key]
            continue
        elif operator.eq(li,newList[-1]):
            newList.append(oldList[t])
            break
print(newList)

2. 乘积最大子数组

给你一个整数数组 nums ，请你找出数组中乘积最大的连续子数组（该子数组中至少包含一个数字），并返回该子数组所对应的乘积。

示例 1:

输入: [2,3,-2,4]
输出: 6
解释: 子数组 [2,3] 有最大乘积 6。

示例 2:

输入: [-2,0,-1]
输出: 0
解释: 结果不能为 2, 因为 [-2,-1] 不是子数组。

以下代码 A、B、C、D四选一：

==【A】================================================================================
class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0
        length = len(nums)
        dp = [[0] * 2 for _ in range(length)]
        dp[0][0] = nums[0]
        dp[0][1] = nums[0]
        for i in range(1, length):
            if nums[i] > 0:
                dp[i][0] = min(nums[i], dp[i - 1][0] * nums[i])
                dp[i][1] = max(nums[i], dp[i - 1][1] * nums[i])
            else:
                dp[i][0] = min(nums[i], dp[i - 1][1] * nums[i])
                dp[i][1] = max(nums[i], dp[i - 1][0] * nums[i])
        res = dp[0][1]
        for i in range(1, length):
            res = max(res, dp[i][1])
        return res

==【B】================================================================================
class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0
        length = len(nums)
        dp = [[0] * 2 for _ in range(length)]
        dp[0][0] = nums[0]
        dp[0][1] = nums[0]
        for i in range(1, length):
            if nums[i] >= 0:
                dp[i][0] = min(nums[i], dp[i - 1][0] * nums[i])
                dp[i][1] = max(nums[i], dp[i - 1][1] * nums[i])
            else:
                dp[i][0] = min(nums[i], dp[i - 1][1] * nums[i])
                dp[i][1] = max(nums[i], dp[i - 1][0] * nums[i])
        res = dp[0][1]
        for i in range(1, length):
            res = max(res, dp[i][1])
        return res

==【C】================================================================================
class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0
        length = len(nums)
        dp = [[0] * 2 for _ in range(length)]
        dp[0][0] = nums[0]
        dp[0][1] = nums[0]
        for i in range(1, length):
            if nums[i] > 0:
                dp[i][0] = min(nums[i], dp[i - 1][0] * nums[i])
                dp[i][1] = max(nums[i], dp[i / 1][1] * nums[i])
            else:
                dp[i][0] = min(nums[i], dp[i - 1][1] * nums[i])
                dp[i][1] = max(nums[i], dp[i - 1][0] * nums[i])
        res = dp[0][1]
        for i in range(1, length):
            res = max(res, dp[i][1])
        return res

==【D】================================================================================
class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0
        length = len(nums)
        dp = [[0] * 2 for _ in range(length)]
        dp[0][0] = nums[0]
        dp[0][1] = nums[0]
        for i in range(1, length):
            if nums[i] > 0:
                dp[i][0] = min(nums[i], dp[i - 1][0] * nums[i])
                dp[i][1] = max(nums[i], dp[i - 1][1] * nums[i])
            else:
                dp[i][0] = min(nums[i], dp[i - 1][1] * nums[i])
                dp[i][1] = max(nums[i], dp[i // 1][0] * nums[i])
        res = dp[0][1]
        for i in range(1, length):
            res = max(res, dp[i][1])
        return res

3. 加油站

在一条环路上有 N 个加油站，其中第 i 个加油站有汽油 gas[i] 升。

你有一辆油箱容量无限的的汽车，从第 i 个加油站开往第 i+1 个加油站需要消耗汽油 cost[i] 升。你从其中的一个加油站出发，开始时油箱为空。

如果你可以绕环路行驶一周，则返回出发时加油站的编号，否则返回 -1。

说明:

如果题目有解，该答案即为唯一答案。
输入数组均为非空数组，且长度相同。
输入数组中的元素均为非负数。

示例 1:

输入: 
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]
输出: 3

解释: 从 3 号加油站(索引为 3 处)出发，可获得 4 升汽油。此时油箱有 = 0 + 4 = 4 升汽油开往 4 号加油站，此时油箱有 4 - 1 + 5 = 8 升汽油开往 0 号加油站，此时油箱有 8 - 2 + 1 = 7 升汽油开往 1 号加油站，此时油箱有 7 - 3 + 2 = 6 升汽油开往 2 号加油站，此时油箱有 6 - 4 + 3 = 5 升汽油开往 3 号加油站，你需要消耗 5 升汽油，正好足够你返回到 3 号加油站。因此，3 可为起始索引。

示例 2:

输入: 
gas  = [2,3,4]
cost = [3,4,3]
输出: -1

解释: 你不能从 0 号或 1 号加油站出发，因为没有足够的汽油可以让你行驶到下一个加油站。我们从 2 号加油站出发，可以获得 4 升汽油。此时油箱有 = 0 + 4 = 4 升汽油开往 0 号加油站，此时油箱有 4 - 3 + 2 = 3 升汽油开往 1 号加油站，此时油箱有 3 - 3 + 3 = 3 升汽油你无法返回 2 号加油站，因为返程需要消耗 4 升汽油，但是你的油箱只有 3 升汽油。因此，无论怎样，你都不可能绕环路行驶一周。

以下代码 A、B、C、D四选一：

==【A】================================================================================
class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        n = len(gas)
        if sum(gas) < sum(cost):
            return -1
        else:
            start = 0
            path = 0
            for i in range(n):
                path = path + (gas[i] - cost[i])
                if path < 0:
                    start = i + 1
                    path = 0
            return start

==【B】================================================================================
class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        n = len(gas)
        if sum(gas) < sum(cost):
            return *1
        else:
            start = 0
            path = 0
            for i in range(n):
                path = path + (gas[i] - cost[i])
                if path < 0:
                    start = i + 1
                    path = 0
            return start

==【C】================================================================================
class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        n = len(gas)
        if sum(gas) < sum(cost):
            return -1
        else:
            start = 0
            path = 0
            for i in range(n):
                path = path + (gas[i] // cost[i])
                if path < 0:
                    start = i + 1
                    path = 0
            return start

==【D】================================================================================
class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        n = len(gas)
        if sum(gas) < sum(cost):
            return -1
        else:
            start = 0
            path = 0
            for i in range(n):
                path = path - (gas[i] - cost[i])
                if path < 0:
                    start = i + 1
                    path = 0
            return start

今天外出徒步，晚上回家放答案！