134. 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 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.
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.
Example 2
Input:
gas = [2,3,4]
cost = [3,4,3]
Output: -1
Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.
题意：给两个数组，第一个数组代表到达当前站点可以补给的 gas， 第二个数组表示为 cost. 问从哪个点起，可以绕其一圈， 注意这个站点是圆形排放的。 首先需要判定是否能够绕一圈。如果总 gas 比总 cost 还少，那么返回 -1。 然后我们从 0 出发， 如果当前的 gas 值大于 cost， 就继续前进。 如果到某一站为负数的话， 就说明从起点到目前站点为止不可能有一个点是起点。 因为一路正着过来，突然变负。 那么从中间的任何一点开始起，正正相加的和都不可能大于从起点之和大。所以起点直接设为下一个点。
代码
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int index = 0;
int ret = 0;
int sum = 0;
for(int i = 0; i < gas.length; i++) {
int tmp = gas[i] - cost[i];
sum += tmp;
ret += tmp;
if (ret < 0) {
index = (i+1)% gas.length;
ret = 0;
}
}
if(sum < 0) return -1;
return index;
}
}