【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 in the clockwise direction, otherwise return -1.
Note:
- If there exists a solution, it is guaranteed to be unique.
- Both input arrays are non-empty and have the same length.
- Each element in the input arrays is a non-negative integer.
N个加油站一个圈，加油gas，耗费cost。
要求选一个加油站出发，可以开车经过所有加油站。
数据特点：具有唯一解，全正数，非空，等长。
【思路】
基于以下两个很机智的结论
1、若第i个加油站，gas[i]-cost[i]<0，必不能从这里出发。
2、因为加油站是一个圈，若ΣΔ(gas[i]-cost[i])<0无解。
【代码】
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int tank=0;//当前站出发后Δ装油量
int start=0;//出发站点
int total=0;//循环总Δ油量
for(int i=0;i<gas.length;i++){
int delta=gas[i]-cost[i];
tank+=delta;
total+=delta;
if(tank<0){
tank=0;
start=i+1;
//站点i出发为负，必然只能从下一站点出发
}
}
return total<0?-1:start;
}
}【举例】
Example 1:
Input:
gas = [1,2,3,4,5]
cost = [3,4,5,1,2] Output: 3 Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.