4.贪心-环形跑道 加油站问题（leetcode 134）

问题描述

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.

翻译

加油站，在一条环形路线上有 N 个加油站，其中第 i 个加油站的油量为 gas[i]。你有一辆油箱容量无限的汽车，从第 i 个加油站到下一个加油站（i+1）需要花费汽油费用[i]。您在其中一个加油站以空油箱开始旅程。如果可以绕行一圈，则返回起始加油站的索引，否则返回-1。

注意：保证解是唯一的。

思路

核心思想：计算前缀和

将每个站的汽油含量-消耗量存放在tem数组中，同时使用currtotal作为总体标量来查看时候汽油数 >路程消耗数 。如果currtotal<0，说明不可能

同时curr作为当前油箱内 油量，如果curr<0，则将curr清0从i+1开始继续遍历

使用while循环

具体证明可以去b站看代码随想录

代码：贪心和暴力

#define  _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string>
#include <algorithm>
#include <stdlib.h>
#include <vector>
using namespace std;
int canCompleteCycleBurte(vector<int> gas, vector<int> cost)
{
	int res = -1;
	int n = gas.size();
	if (n == 1 && gas[0] >= cost[0])
		return 0;
	for (int i = 0; i < n; i++)
	{
		int curr = 0;
		
		if (gas[i] == cost[i])
			continue;
		int j = i;
		do
		{
			if (curr + gas[j % n] >= cost[j % n])
			{
				curr += (gas[j % n] - cost[j % n]);
				j++;
				if (j == n + i)
					return i;
			}
			else
			{
				j++;
				break;
			}

		} while (j % n != i);
	}
	return res;
}
int canCompleteCycleGreedy(vector<int>gas, vector<int>cost)
{
	int n = gas.size();
	int total = 0;
	int res = 0; int curr = 0;
	for (int i = 0; i < n; i++)
	{
		total += gas[i] - cost[i];//总油数-总耗油
		curr += gas[i] - cost[i];//遍历到当前的车站，油箱剩余油量
		if (curr < 0)
		{
			res = i + 1;//说明前面任意一个车站出发都不能到达第i个，因此从i+1开始
			curr = 0;
		}
	}
	if (total < 0)//总油数少于总耗油
		return -1;
	else
		return res;
}

int main()
{
	
	int n;
	cout << "Please enter the number of the gas station :" << endl;
	cin >> n;
	vector<int>gas(n),cost(n);
	cout << "Please enter the gas that each station has :" <<  endl;
	for (int i = 0; i < n; i++)
		cin >> gas[i];
	cout << "Please enter the cost of gas go to the next station :" << endl;
	for (int i = 0; i < n; i++)
		cin >> cost[i];
	//int res = canCompleteCycleBurte(gas, cost);
	int res = canCompleteCycleGreedy(gas, cost);

	if (res == -1)
		cout << "There is no way to complete ! " << res << endl;
	else
		cout << "We should begin at " << res << " station " << endl;
	return 0;
}

测试案例

输入: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
输出: 3
解释:
从 3 号加油站(索引为 3 处)出发，可获得 4 升汽油。此时油箱有 = 0 + 4 = 4 升汽油
开往 4 号加油站，此时油箱有 4 - 1 + 5 = 8 升汽油
开往 0 号加油站，此时油箱有 8 - 2 + 1 = 7 升汽油
开往 1 号加油站，此时油箱有 7 - 3 + 2 = 6 升汽油
开往 2 号加油站，此时油箱有 6 - 4 + 3 = 5 升汽油
开往 3 号加油站，你需要消耗 5 升汽油，正好足够你返回到 3 号加油站。
因此，3 可为起始索引。

运行结果

时间复杂度分析

Only use one loop toIterate over the array cost O(n);

速记