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134. Gas Station

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134. Gas Station

题目

 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. 

解析

  • 思路1:可以穷举每个位置,求和是否大于0;
  • 非常经典的一道题。可以转换成求最大连续和做,但是有更简单的方法。
  • 基于一个数学定理:如果一个数组的总和非负,那么一定可以找到一个起始位置,从他开始绕数组一圈,累加和一直都是非负的
  • 继续对问题进行抽象,我们肯定希望一开始的路程都是加油>消耗的,这样就可以累积油量应付后面的消耗量大的,根据这个思路,就转变成求数组的最大连续子序列的和(并且记录起始下标),这是我们前面接触过的题目。
  • 因为是循环数组,我们还得考虑一种情况:就是首尾都是正数,怎么办呢?
  • 正因为是循环数组,还有一种类似于快排前后指针的操作。
class Solution {
public:
	int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {

		if (gas.size()==0||cost.size()==0||gas.size()!=cost.size())
		{
			return -1;
		}
		int index = 0;
		int sum_resdual = 0;

		for (int j = 0; j < gas.size();j++)
		{
			sum_resdual += (gas[j] - cost[j]);
		}
		if (sum_resdual<0)
		{
			return -1;
		}

		sum_resdual = 0;
		for (int i = 0; i < gas.size();i++)
		{
			//index = 0;  //bug
			sum_resdual += (gas[i] - cost[i]);
			if (sum_resdual<0)
			{
				index = i + 1;
				sum_resdual = 0;
			}
		}

		return index;
	}
};

  • 换个写法
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
  int start = 0; // 起始位置
  int remain = 0; // 当前剩余燃料
  int debt = 0; // 前面没能走完的路上欠的债

  for (int i = 0; i < gas.size(); i++) {
    remain += gas[i] - cost[i];
    if (remain < 0) {
      debt += remain;
      start = i + 1;
      remain = 0;
    }
  }

  return remain + debt >= 0 ? start : -1;  //这句话保证了求和sum<0,返回-1
}
  • 从start出发, 如果油量足够, 可以一直向后走 end++; 油量不够的时候,start向后退 最终 start == end的时候,如果有解一定是当前 start所在位置

class Solution {
public:  int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {      
        int start = gas.size() - 1;
        int end = 0;
        int sum = gas[start] - cost[start];
        while(start > end){
            if(sum >= 0){
                sum += gas[end] - cost[end];
                ++end;
            }else{
                --start;
                sum += gas[start] - cost[start];
            }
        }
        return sum >=0 ? start : -1;
         
         
    }
};

题目来源

posted @ 2017-12-27 18:48  ranjiewen  阅读(226)  评论(0编辑  收藏  举报