代码随想录算法训练营第34天|● 1005.K次取反后最大化的数组和 ● 134. 加油站● 135. 分发糖果

最新推荐文章于 2025-12-03 17:03:32 发布

原创最新推荐文章于 2025-12-03 17:03:32 发布 · 144 阅读

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#算法

文章包含三个编程问题的解决方案：1)给定一个数组和K次取反操作，如何最大化数组和；2)在加油站的路径规划问题，判断汽车能否完成一圈；3)分发糖果问题，确保评价高的人至少获得的糖果数不少于评价低的人。每个问题都涉及数组处理和优化策略。

1005.K次取反后最大化的数组和

代码随想录

class Solution(object):
    def largestSumAfterKNegations(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        index=0
        #flip the samllest index every time
        for i in range(k):
            #find the smallest index
            for j in range(len(nums)):
                if nums[j]<=nums[index]:
                    index=j
            nums[index]=-nums[index]
        return sum(nums)

134. 加油站

代码随想录

class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        #start_index
        start = 0
        #current sum of gas
        curSum = 0
        #total sum of gas
        totalSum = 0
        for i in range(len(gas)):
            curSum += gas[i] - cost[i]
            totalSum += gas[i] - cost[i]
            #if current usm of gas is smaller than 0, use next stop as start. 
            if curSum < 0:
                curSum = 0
                start = i + 1
        #if total sum of gas is smaller than 0, the car must not reach goal 
        if totalSum < 0: return -1
        #else we can return the result
        return start

135. 分发糖果

代码随想录

class Solution(object):
    def candy(self, ratings):
        """
        :type ratings: List[int]
        :rtype: int
        """
        #candy to give
        candyVec = [1] * len(ratings)
        #loop from left
        for i in range(1, len(ratings)):
            #if ranking of right is higher, give one more candy 
            if ratings[i] > ratings[i - 1]:
                candyVec[i] = candyVec[i - 1] + 1
        #loop from right
        for j in range(len(ratings) - 2, -1, -1):
            #if ranking of left is higher, give one more candy 
            if ratings[j] > ratings[j + 1]:
                candyVec[j] = max(candyVec[j], candyVec[j + 1] + 1)
        #we have a complete result
        return sum(candyVec)