134. Gas Station (Array; DP)

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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：

1. 是否能travel around? =>只要gas[]加总的和 - cost[]加总的和 > 0，就能够。=>需要一个INT计算gas[]-cost[]的总合

2. 如何找到起点？如果到了某一站油不够，那么起点必须从它后面开始重新找。=>需要一个INT记录从找到的起点到目前为止的油量

时间复杂度O(n)

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int sum = 0, total = 0, len = gas.size(), index = -1;
        for(int i=0; i<len; i++){
            sum += gas[i]-cost[i];
            total += gas[i]-cost[i];
            if(sum < 0){
                index = i;
                sum = 0;
            }
        }
        return total>=0 ? index+1 : -1;
    }
};

134. Gas Station (Array; DP)

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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4861262.html