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gas_station——leetcode

方法1:

1. 每一站的代价为gas-cost, 也就是求从哪一站开始累加代价和总是大于0。

2. 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。

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        // Note: The Solution object is instantiated only once.
		int total = 0;
		int currentgas = 0; 
		int startpoint = -1;
		int sz = gas.size();
		for(int i = 0; i < sz; i++)
		{
			currentgas += gas[i] - cost[i];
			total += gas[i] - cost[i];
			if(currentgas < 0)
			{
				startpoint = i;
				currentgas = 0;
			}
		}
		return total >= 0 ? startpoint+1 : -1;
    }
};


方法2:

class Solution {
public:
	int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
		int len = gas.size();
		int tank_all = 0;
		int pos = 0;
		int tank_cur =0;
		for (int i = 0; i < len; i++){
			int tank_remain = gas[i] - cost[i];
			if (tank_cur >= 0){
				tank_cur += tank_remain;
			}
			else{//find a  new start .restart
				tank_cur = tank_remain;
				pos = i;
			}
			tank_all += tank_remain;
		}
		return tank_all >= 0 ? pos : -1;
	}
};

gas_station——leetcode