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.

Greedy

SOLUTION 1:

引自Lexi‘s Leetcode solutions 的解答:

1. 从i开始，j是当前station的指针，sum += gas[j] – cost[j] （从j站加了油，再算上从i开始走到j剩的油，走到j+1站还能剩下多少油）
2. 如果sum < 0，说明从i开始是不行的。那能不能从i..j中间的某个位置开始呢？既然i出发到i+1是可行的， 又i~j是不可行的， 从而发现i+1~ j是不可行的。
3. 以此类推i+2~j， i+3~j，i+4~j 。。。。等等都是不可行的
4. 所以一旦sum<0，index就赋成j + 1，sum归零。
5. 最后total表示能不能走一圈。

以上做法，其实是贪心的思想：

也就是说brute force的解是 ： 一个一个来考虑， 每一个绕一圈， 但是 实际上 我们发现 i - j不可行 直接index就跳到j+1， 这样周而复始，很快就是绕一圈 就得到解了。

 1 public class Solution { 2     public int canCompleteCircuit(int[] gas, int[] cost) { 3         if (gas == null || cost == null || gas.length == 0 || cost.length == 0) { 4             // Bug 0: should not return false; 5             return -1; 6         } 7          8         int total = 0; 9         int sum = 0;10         11         int startIndex = 0;12         13         int len = gas.length;14         for (int i = 0; i < len; i++) {15             int dif = gas[i] - cost[i];16             sum += dif;17             18             if (sum < 0) {19                 // Means that from 0 to this gas station, none of them can be the solution.20                 startIndex = i + 1; // Begin from the next station.21                 sum = 0; // reset the sum.22             }23             24             total += dif;25         }26         27         if (total < 0) {28             return -1;29         }30         31         return startIndex;32     }33 }

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/greedy/CanCompleteCircuit.java

LeetCode: Gas Station 解题报告