LeetCode : 134. Gas Station加油

冷不防 2021-06-12 20:36 375阅读 0赞

**试题**  
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.

**代码**

class Solution {
        public int canCompleteCircuit(int[] gas, int[] cost) {
            int[] need = new int[gas.length];
            for (int i = 0; i < gas.length; i++) {
                need[i] = gas[i] - cost[i];
            }
            
            int sum = 0;
            for (int i = 0; i < need.length; i++) {
                sum += need[i];
            }
            if (sum < 0) return -1;
            
    		// 输出的index肯定是从前面是负数，后面index是正数或0开始；那么从这个时候开始sum统计，如果sum大于等于0显然可以继续往下统计，如果小于0显然从这个index开始是无法到达尾部。最后一种情况是这个sum横跨了need的尾部和头部，刚好是最后一个index。
            int start = 0;
            sum = 0;
            for (int i = 0; i < need.length; i++) {
                sum += need[i];
                if (sum < 0) {
                    start = i + 1;
                    sum = 0;
                } 
            }
            return start;
        }
    }