leetcode_134_Gas Station

标签：array   c++   leetcode   

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.

Note:
The solution is guaranteed to be unique.

//方法一：O( N) 的解法是，设置两个变量，sum表示起点为station时当前的总油量；total则为全程的总油量，如果total不小于0，则返回通过sum得到起点的下标station；如果total小于0，则返回 -1。
//时间复杂度 O(n) ，空间复杂度 O(1)
class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int total = 0;
        int station = 0;
        for(int i = 0, sum = 0; i < gas.size(); i++)
        {
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];
            if(sum < 0)
            {
                station = i + 1; //起点位置
                sum = 0;
            }
        }
        return total < 0 ? -1 : station;
    }
};

//方法：超时， O (N^2) 的解法，对每个点进行模拟
class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int sum = 0;
        for(int i = 0; i < gas.size(); i++)
        {
            sum = 0;
            for(int j = i; j < gas.size(); j++)
            {
                sum += gas[j] - cost[j];
                if(sum < 0)
                    break;
            }
            if(sum < 0)
                continue;
            for(int j = 0; j < i; j++)
            {
                sum += gas[j] - cost[j];
                if(sum < 0)
                    break;
            }
            return i;
        }
        return -1;
    }
};

leetcode_134_Gas Station

标签：array   c++   leetcode   

原文地址：http://blog.csdn.net/keyyuanxin/article/details/46591715