原题传送门

1. 题目描述

2. Solution 1

1、思路分析
以gas=[1,2,3,4,5], cost=[3,4,5,1,2]为例
gas: 1 2 3 4 5
cost: 3 4 5 1 2
tank: -2 -2 -2 3 3

Step 1: 先分析是否有解。对上面的tank求和为total=0，当total>=0时有解。
Step 2: 若无解直接返回-1；在有解的情况下，分析找出起点start。第一个tank>=0的下标即为起点。

2、代码实现

package Q0199.Q0134GasStation;

/*
    1> If car starts at A and can not reach B. Any station between A and B
    can not reach B.(B is the first station that A can not reach.)
    2> If the total number of gas is bigger than the total number of cost. There must be a solution.
 */
public class Solution {

    public int canCompleteCircuit(int[] gas, int[] cost) {
        int start = 0, total = 0, tank = 0;
        // if car fails at start, record the next station
        for (int i = 0; i < gas.length; i++)
            if ((tank = tank + gas[i] - cost[i]) < 0) {
                start = i + 1;
                total += tank;
                tank = 0;
            }
        return (total + tank < 0) ? -1 : start;
    }
}

/*
If you are confused,(like I was before),try think about it with two passes. Use the first pass to determine if we
have a solution(property 2 above). Then use the second pass to find out the start position(use property 1).
After you are comfortable with 2 passes, you can absolutely modify it into one pass solution.

    public int canCompleteCircuit(int[] gas, int[] cost) {
        //determine if we have a solution
        int total = 0;
        for (int i = 0; i < gas.length; i++) {
            total += gas[i] - cost[i];
        }
        if (total < 0) {
            return -1;
        }

        // find out where to start
        int tank = 0;
        int start = 0;
        for (int i = 0; i < gas.length;i++) {
            tank += gas[i] - cost[i];
            if (tank < 0) {
                start = i + 1;
                tank = 0;
            }
        }
        return start;
    }
 */

3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(1)

标签：0134,tank,int,gas,Gas,start,Station,total,cost
来源： https://www.cnblogs.com/junstat/p/16294353.html