原题传送门->134. Gas Station

Description

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station’s index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

Example 1:

Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.

Example 2:

Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can’t start at station 0 or 1, as there is not enough gas to travel to the next station.
Let’s start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can’t travel around the circuit once no matter where you start.

Constraints:

gas.length == n
cost.length == n
1 <= n <= 10^4
0 <= gas[i], cost[i] <= 10^4

法1：暴力：

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        for(int i = 0; i < gas.size(); ++i){
            int acc_gas = gas[i];
            int acc_cost = cost[i];
            if(acc_gas < acc_cost) continue;
            int flag = 1;
            for(int j = (i+1)%gas.size(); j != i; j = ((j+1)%gas.size()) ){
                acc_cost += cost[j];
                acc_gas += gas[j];
                if(acc_gas < acc_cost){
                    flag = 0;
                    break;
                }
            }

            if(flag) return i;
        }
        return -1;
    }
};

法2：
参考折线图分析

大概思想：

假设让车从0号位开始跑一遍，就能得到在各个加油站剩余油量。
这样就得到了一张折线图。纵坐标为剩余油量。

倘若改变起点，这张折线图的形状不会改变，仅仅是做上下平移。

只要让所有的点平移到坐标轴上方就OK了。

把起点放在哪儿呢？设折线图的最低点为i,起点就是i的后一个位置。

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int acc = 0, acc_min = 10, ans = 0;
        for(int i = 0; i < gas.size(); ++i){
            acc += (gas[i] - cost[i]);
            if(acc < acc_min){
                ans = i;
                acc_min = acc;
            }            
        }
        if(acc < 0)
        return -1;
        return (ans+1) % gas.size();
    }
};