[LeetCode] 134. Gas Station

Topic links: https://leetcode-cn.com/problems/gas-station/

Subject description:

There are N stations in a loop, wherein the i-th gasoline stations Gas [i] l.

You have an unlimited capacity of the fuel tank car, from the i-th gas station bound for the first i + 1 gas stations need to consume gasoline cost [i] liter. You start from a gas station where, at the start of the fuel tank is empty.

If you can travel around the ring a week starting when the number of gas stations is returned, otherwise -1.

Description:

• If the title has a solution, the answer is the only answer.
• Input array are non-empty array, and the same length.
• Enter the array elements are non-negative.

Example:

Example 1:

输入: 
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]

输出: 3

解释:
从 3 号加油站(索引为 3 处)出发，可获得 4 升汽油。此时油箱有 = 0 + 4 = 4 升汽油
开往 4 号加油站，此时油箱有 4 - 1 + 5 = 8 升汽油
开往 0 号加油站，此时油箱有 8 - 2 + 1 = 7 升汽油
开往 1 号加油站，此时油箱有 7 - 3 + 2 = 6 升汽油
开往 2 号加油站，此时油箱有 6 - 4 + 3 = 5 升汽油
开往 3 号加油站，你需要消耗 5 升汽油，正好足够你返回到 3 号加油站。
因此，3 可为起始索引。

Example 2:

输入: 
gas  = [2,3,4]
cost = [3,4,3]

输出: -1

解释:
你不能从 0 号或 1 号加油站出发，因为没有足够的汽油可以让你行驶到下一个加油站。
我们从 2 号加油站出发，可以获得 4 升汽油。 此时油箱有 = 0 + 4 = 4 升汽油
开往 0 号加油站，此时油箱有 4 - 3 + 2 = 3 升汽油
开往 1 号加油站，此时油箱有 3 - 3 + 3 = 3 升汽油
你无法返回 2 号加油站，因为返程需要消耗 4 升汽油，但是你的油箱只有 3 升汽油。
因此，无论怎样，你都不可能绕环路行驶一周。

Ideas:

From ithe jposition, there is sum(gas) < sum(cost)described ito not j, and ito jbetween any locations are not toj

You can see proof of official explanations

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        if sum(gas) < sum(cost):return -1
        res = 0
        tank = 0
        for i, item in enumerate(zip(gas, cost)):
            tank += (item[0] - item[1])
            if tank < 0:
                res = i + 1
                tank = 0
        return res

java

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int sum = 0;
        for (int i = 0; i < gas.length; i++) sum += (gas[i] - cost[i]);
        if (sum < 0) return -1;
        int tank = 0;
        int res = 0;
        for (int i = 0; i < gas.length; i++) {
            tank += (gas[i] - cost[i]);
            if (tank < 0) {
                res = i + 1;
                tank = 0;
            }
        }
        return res;
    }
}