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.

之前自己想的解决方案是O(n^2)的，类似于DFS的方案，百度之，发现一个O(n)的解决方案，地址（http://blog.csdn.net/jellyyin/article/details/12245429）。

解决思路简要如下：

1. 首先假设gas[i]-cost[i] 的差的数组为 diff[i]

2. 全局有解的重要条件是 diff 数组的和大于等于0

3. 从零往后遍历diff数组，记录两个值，一个全局的和，一个局部的和，两个和的算法是一样的，但用处不一样， 全局的和用来最后判断是否有解，局部的和稍微麻烦一点，我们用从0开始得局部和举例子，假如说在 k 碰到第一个小于零的局部和，那么起点不可能在 0~k 之间的任意一点。这时假设起点为 (k+1)%diff.length, 然后清零局部和。

代码如下：

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int length = gas.length;
        int[] diff = new int[length];
        for(int i=0;i<length;i++){
            diff[i] = gas[i] - cost[i];
        }
        int start = 0;
        int sum = 0;
        int part_sum =0;
          
        for(int i=0;i<length;i++){
            sum += diff[i];
            part_sum += diff[i];
              
            if(part_sum<0){
                start = (i+1)%length;
                part_sum = 0;
            }
        }
        return sum>=0?start:-1;
    }
}