【題目】
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, otherwise return -1.
Note:
The solution is guaranteed to be unique.
【思路】
Greedy algorithm..之前有見過這道題。
環形的station. 找到能夠滿足走完全程的起始點。
在某一站的消耗 : gas[i] - cost[i]
剩下的能源: sum = sum + ( gas [i] - cost [i] );只要在中途出現 sum < 0那麼這個點就失敗了。
第一種方法,brute force. 每一個點都走一次, n*n = n^2.
第二種方法:
開始 sum = 0;
start是最後一個,end是最前面一個,
【代碼】
int start = gas.size()-1;
int end = 0;
int sum = gas[start] - cost[start];
while (start > end) {
if (sum >= 0) {
sum += gas[end] - cost[end];
++end;
}
else {
--start;
sum += gas[start] - cost[start];
}
}
return sum >= 0 ? start : -1;
O(n)
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int sum = 0;
int total = 0;
int j = -1;
for (int i = 0; i < gas.length; i++) {
sum += gas[i] - cost[i];
total += gas[i] - cost[i];
if(sum < 0) { //之前的油量不夠到達當前加油站
j = i;
sum = 0;
}
}
if (total < 0) return -1; //所有加油站的油量都不夠整個路程的消耗
else return j + 1;
}
}
Brute Force
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
//依次從每一個加油站出發
for (int i = 0; i < gas.length; i++) {
int j = i;
int curgas = gas[j];
while (curgas >= cost[j]) { //如果當前的汽油量能夠到達下一站
curgas -= cost[j]; //減去本次的消耗
j = <span style="background-color: rgb(255, 255, 153);">(j + 1) % gas.length; </span>//到達下一站
if (j == i) return i; //如果回到了起始站,那麼旅行成功
curgas += gas[j]; //到下一站後重新加油,繼續前進
}
}
return -1;
}
}