LeetCode _ Gas Station

最新推荐文章于 2021-02-15 21:03:58 发布

Surenon

最新推荐文章于 2021-02-15 21:03:58 发布

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原文链接：http://www.cnblogs.com/graph/p/3352042.html

本文介绍了一个算法问题，即如何在一个环形路线上找到一个起点，使得一辆油箱无限大的汽车能够通过合理的加油策略完成一次完整的环路旅行。该问题通过一个不回溯的策略解决，确保了从选定的起点出发，汽车总能成功回到起点。

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.

思路：类似KMP不回朔的思想。start表示开始的点，sum表示当前汽车的油量。当汽车到达汽油站i时如果不能到达下一站，则更新start直到可以使汽车能够从当前节点到达下一站，如果不存在，则把start设置为下一站。

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
       int len = gas.size();
       if(cost.size() != len) return -1;
       vector<int> flag(len*2, 0);
       for(int i = 0; i< len; i++)
            flag[i] = gas[i] - cost[i];
       for(int i = len ; i< len *2; ++i)
            flag[i] = flag[i-len];
    
       int start = 0 , sum = 0;         
       for(int i = 0; i< len + start && start < len; )
        {
            sum += flag[i] ;
            if(sum >= 0){
                ++i;
                continue;
            } 
            while(sum< 0 && start < i){
                sum -= flag[start];
                ++start;
            }
            i++;
            if(sum < 0){
                sum = 0;
                start = i;
            }
            
        }
        if(start <len)
         return start;
        return -1;
        
    }
};