LeetCode 134. Gas Station

原题链接在这里：https://leetcode.com/problems/gas-station/

题目：

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.

题解：

sumCur是走到当前油站的剩余油量, totalCir是走完一整圈的剩余油量。若是从起始站走到i站，当前剩余油量sumCur < 0说明走不到i站，同时也不能走到起始站到i站中间的任意一站。

为什么走不到起点到i 战中的任意一战呢. 假设k站是起始站到i站中间的一站，因为是走到i才停的，走到k并没有停，所以走到k时sumCur还是正数，但走到i时sumCur变成了负数。

现在假设从k站其实那么sumCur是0, 走到i的话肯定是一个负数，因为正数的sumCur走到i都被减成了负数，何况是0的sumCur呢.

totalCir 其实是用来判定能不能走完一圈，若是totalCir是正数，那说明从i点能走完一圈，从i开始的一圈和从0开始的一圈用油不变，油量不变，所以可以直接使用。直接返回index+1.

Note: 注意一种情况就是从0站开始的SumCir一直是非负的话， 返回index+1, 此时和index的初始化有直接关系. index初始化为-1. index+1正好是0.

Time Complexity: O(n). n = gas.length.

Space: O(1). 

AC Java:

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        if(gas == null || cost == null || gas.length == 0 || cost.length == 0 || gas.length != cost.length){
            return -1;
        }
        int sumCur = 0;     //到达当前油站的剩余油量
        int totalCir = 0;   //走完一整圈的剩余油量
        int index = -1;
        for(int i = 0; i<gas.length; i++){
            int diff = gas[i] - cost[i];
            sumCur += diff;
            totalCir += diff;
            if(sumCur < 0){ //走到i站是负数，说明走不到i站，也不能选取起始站到i站中的任何一战
                sumCur = 0;
                index = i;  //更新index是和 index的初始条件有关系的
            }
        }
        return totalCir >= 0 ? index+1:-1;
    }
}