There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an e…

There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an e…

https://leetcode.com/problems/gas-station/ 题目: There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next…

There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an e…

There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an e…

  There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an…

原题 题意: 过一个循环的加油站,每个加油站可以加一定数量的油,走到下一个加油站需要消耗一定数量的油,判断能否走一圈. 思路: 一开始思路就是遍历一圈,最直接的思路. class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int beg = 0; int tank = 0; int n = gas.size(); int sumGas = 0, s…

134. Gas Station 不会做. 1. 朴素的想法,就是针对每个位置判断一下,然后返回合法的位置,复杂度O(n^2),显然会超时. 把这道题转化一下吧,求哪些加油站不能走完一圈回到自己,要求O(N)的复杂度. 如果sgas < scost,那么显然所有的站点都无法走完一圈. 2. 考虑怎么进行化简,寻找有没有什么可以利用的性质,考虑可不可以进行递推,对每个位置求出gas - cost,然后很明显观察到如果一个位置为负数,那么这个位置显然不能走完一圈,那么接下来 考虑怎么进行简化,找到负…

870. Advantage Shuffle 思路:A数组的最大值大于B的最大值,就拿这个A跟B比较:如果不大于,就拿最小值跟B比较 A可以改变顺序,但B的顺序不能改变,只能通过容器来获得由大到小的顺序,并且必须存储相应的index,因为最终需要将选择的A的数值存入与这个B相对应的index下 class Solution { public: vector<int> advantageCount(vector<int>& A, vector<int>&…

[LeetCode]Gas Station 解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/gas-station/#/description 题目描述: There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tan…