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.
这个题最直观的解法就是从某个点开始绕一圈判断是否可以走完整圈。

public class Solution {
	public int canCompleteCircuit(int[] gas, int[] cost) {
		for(int i=0;i<gas.length;i++){
			if(canComplete(i, gas, cost)){
				return i;
			}
		}
		return -1;
	}
	// 每一点开始循环一圈
	boolean canComplete(int start,int[] gas, int[] cost){
		int oldStart = start;
		int oilcount = 0;
		int len = gas.length;
		while(true){
			if(start>=len && start % len == oldStart){
				break;
			}
			oilcount += gas[start%len];
			if(oilcount<cost[start%len]){
				return false;
			}
			oilcount -= cost[start%len];
			start++;
		}
		return true;
	} 
}

但是注意到题目说到肯定有一个答案，所以可以进行时间上的优化。上面的时间复杂度O（n^2），现在优化时间复杂度为O（n）。
思路：如果所有站的代价和大于0，则所求的路线必定存在。如果总代价〉=0,从序号0开始求代价和，如果代价和小于0，则不是从本站或者本站之前的某一个代价大于0的站开始，必从下一站即之后的站开始，而且这样的站必定存在，时间复杂度O（n）。题目没有提示不存在返回什么，判题系统不存在为-1。

class Solution {

public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int sum=0;
        int total=0;
        int start=0;
        for(int i=0;i<gas.size();i++)
        {
            sum+=gas[i]-cost[i];
            total+=gas[i]-cost[i];
            if(sum < 0)
            {
                start=(i+1)%gas.size(); sum=0;
            }
        }
        
        if(total <0)
            return -1;
        else
            return start;
    }

};