leetCode---Gas Station
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一. 题目:Gas Station
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 empty tank at one of the gas stations.
Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.
二. 思路分析
题目大意:环形路线上有N个加油站,每个加油站有汽油gas[i],从每个加油站到下一站消耗汽油cost[i],问从哪个加油站出发能够回到
起始点,如果都不能则返回-1。思路分析:我们首先要知道能走完整个环的前提是gas的总量要大于cost的总量,这样才会有起点的存在。
假设开始设置起点start = 0, 并从这里出发,如果当前的gas值大于cost值,就可以继续前进,此时到下一个站点,剩余的gas加上当前的gas再
减去cost,看是否大于0,若大于0,则继续前进。当到达某一站点时,若这个值小于0了,则说明从起点到这个点中间的任何一个点都不能作为
起点,则把起点设为下一个点,继续遍历。当遍历完整个环时,当前保存的起点即为所求。代码如下:
class Solution {public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int total = 0, sum = 0, start = 0; for (int i = 0; i < gas.size(); ++i) { total += gas[i] - cost[i]; sum += gas[i] - cost[i]; if (sum < 0) { start = i + 1; sum = 0; } } if (total < 0) return -1; else return start; }};
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