Gas Station
題意:
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.
Example
Given 4 gas stations with gas[i]=[1,1,3,1]
, and the cost[i]=[2,2,1,1]
. The starting gas station's index is 2
.
Note The solution is guaranteed to be unique.
Challenge O(n) time and O(1) extra space
解題思路:
由於此題只需要找出是否有可能的解,並回傳起始的index,因此我們只需要檢查所有 diff是否為正,若為正,則回傳該 startIndex。
引自 Lexi 網友的解答:
- 從i開始,j是當前station的指針,sum += gas[j] – cost[j] (從j站加了油,再算上從i開始走到j剩的油,走到j+1站還能剩下多少油)
- 如果sum < 0,說明從i開始是不行的。那能不能從i..j中間的某個位置開始呢?既然i出發到i+1是可行的, 又i~j是不可行的, 從而發現i+1~ j是不可行的。
- 以此類推i+2~j, i+3~j,i+4~j 。。。。等等都是不可行的
- 所以一旦sum<0,index就賦成j + 1,sum歸零。
- 最後total表示能不能走一圈。
public class Solution {
/**
* @param gas: an array of integers
* @param cost: an array of integers
* @return: an integer
*/
public int canCompleteCircuit(int[] gas, int[] cost) {
if (gas == null || gas.length == 0 || cost == null || cost.length == 0) {
return -1;
}
int len = cost.length;
int total = 0; // 累積整個環的所有差值,如果結果為負的話,代表找不到任何結果
int sum = 0; // 指從j站加了油,再算上從i開始走到j剩的油,走到j+1站還能剩下多少油
int startIndex = 0; // 指從哪裡開始走
for (int i = 0; i < len; i++) {
int diff = gas[i] - cost[i];
sum += diff;
if (sum < 0) {
// 表示從第0個加油站到目前這個加油站沒有解答
// 因此從下一個加油站開始繼續找。
startIndex = i + 1;
sum = 0;
}
total += diff;
}
// total是否為正,若為負則表示沒有任何答案,返回-1,否則返回startindex
if (total < 0) {
return -1;
} else {
return startIndex;
}
}
}