【LeetCode】134. Gas Station

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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 in the clockwise direction, otherwise return -1.

Note:

• If there exists a solution, it is guaranteed to be unique.
• Both input arrays are non-empty and have the same length.
• Each element in the input arrays is a non-negative integer.

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.

链接：https://leetcode.com/problems/gas-station/description/

题解：本题比较难，主要入手角度需要正确，首先本题保证了只有一个满足条件的节点或者不存在，那么我们从任意一个起点开始遍历，并且每次遍历记录当前油量，需要发现如果第一次遇到j不能到达那么之前的点都不能到达，并把起点更新为下一个点，这是很容易证明的因为如果中间点m能到j，那么i肯定能到j因为i能到m所以你需要的变量就是tank通过判断是否大于o以及start标记目标节点，并且tank重置为0，还有tot判断是否存在目标节点，代码如下：

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