【Leetcode】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.

【思路】

Greedy algorithm..之前有见过这道题。

环形的station. 找到能够满足走完全程的起始点。

在某一站的消耗 ： gas[i] - cost[i]

剩下的能源： sum = sum + ( gas [i] - cost [i] );只要在中途出现 sum < 0那么这个点就失败了。

第一种方法，brute force. 每一个点都走一次， n*n  = n^2.

第二种方法：

开始 sum = 0;

start是最后一个，end是最前面一个，

【代码】

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;

O(n)

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int sum = 0;
        int total = 0;
        int j = -1;
        for (int i = 0; i < gas.length; i++) {
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];
            if(sum < 0) {   //之前的油量不够到达当前加油站
                j = i;
                sum = 0;
            }
        }
        if (total < 0) return -1;    //所有加油站的油量都不够整个路程的消耗
        else return j + 1;
    }
}

Brute Force

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        //依次从每一个加油站出发
        for (int i = 0; i < gas.length; i++) {
            int j = i;
            int curgas = gas[j];
            
            while (curgas >= cost[j]) { //如果当前的汽油量能够到达下一站
                curgas -= cost[j];      //减去本次的消耗
                
                j = <span style="background-color: rgb(255, 255, 153);">(j + 1) % gas.length; </span>//到达下一站
                if (j == i) return i;   //如果回到了起始站，那么旅行成功
                
                curgas += gas[j];       //到下一站后重新加油，继续前进
            }
        }
        return -1;
    }
}