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.

之前自己想的解決方案是O(n^2)的，類似於DFS的方案，百度之，發現一個O(n)的解決方案，地址（http://blog.csdn.net/jellyyin/article/details/12245429）。

解決思路簡要如下：

1. 首先假設gas[i]-cost[i] 的差的數組爲 diff[i]

2. 全局有解的重要條件是 diff 數組的和大於等於0

3. 從零往後遍歷diff數組，記錄兩個值，一個全局的和，一個局部的和，兩個和的算法是一樣的，但用處不一樣， 全局的和用來最後判斷是否有解，局部的和稍微麻煩一點，我們用從0開始得局部和舉例子，假如說在 k 碰到第一個小於零的局部和，那麼起點不可能在 0~k 之間的任意一點。這時假設起點爲 (k+1)%diff.length, 然後清零局部和。

代碼如下：

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;
    }
}