LeetCode134: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 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.
Note:
The solution is guaranteed to be unique.
解题思路:
一开始,我打算通过两层循环,依次从某个点出发并测试是否能够运行一圈,可想而知时间复杂度为O(n2),不满足要求。之后看了http://blog.csdn.net/kenden23/article/details/14106137这篇博客的解题思路,才发现有更简单更优雅的解决方案,大概思路如下:
1 如果总的gas - cost小于零的话,那么没有解返回-1
2 如果前面所有的gas - cost加起来小于零,那么前面所有的点都不能作为出发点。
关于第一点无需多言,这里详解下第二点,为什么前面所有的点都不能作为起始站了,原因是:
假设从第0个站点开始,0~1,剩余的煤气left1 = gas[i]-cost[i],如果left为负,则过不去,必须从下一个站点从新开始,如果为正,则1~2时,left2 = gas[1]+left – cost[1],然后是2~3等等继续下去,如果left一直为正,则表示这些站点都可以过去,但当某个站点i~i+1时,left为负数,说明过不去,且之前的所有站点都不能作为起始站,因为,每个站点要到下一个站点时,gas = gas +left,都不能过去,现在如果从某个站点开始,即gas量为它自身,更过不去。
实现代码:
#include <iostream>
#include <vector>
using namespace std; /*
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. Note:
The solution is guaranteed to be unique.
*/
class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
if(gas.size() == || cost.size() == || gas.size() != cost.size())
return -;
int left = ;
int total = ;
int start = ;
int n = gas.size();
for(int i = ; i < n; i++)
{
left += gas[i] - cost[i];//从i到i+i,剩余的煤气
total += gas[i] - cost[i];
if(left < )//表示前面那些站点都不能作为起始站,现在开始从下一个站点开始
{
start = i + ;
left = ;
}
}
return total >= ? start : -;//煤气总量是否大于等于总消耗 }
};
ing main(void)
{
return ;
}
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