Gas Station

描述

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.

分析

首先想到的是O(N2)O(N^2)的解法，对每个点进行模拟。

O(N)的解法是，设置两个变量，sum判断当前的指针的有效性；total则判断整个数组是否有解，有就返回通过sum得到的下标，没有则返回-1。

代码

  1. // Gas Station
  2. // 时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4. public:
  5.     int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
  6.         int total = 0;
  7.         int j = -1;
  8.         for (int i = 0, sum = 0; i < gas.size(); ++i) {
  9.             sum += gas[i] - cost[i];
  10.             total += gas[i] - cost[i];
  11.             if (sum < 0) {
  12.                 j = i;
  13.                 sum = 0;
  14.             }
  15.         }
  16.         return total >= 0 ? j + 1 : -1;
  17.     }
  18. };

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/array/gas-station.html