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.
1、暴力的做法
public class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int len = gas.length; int sum = 0; int[] dp = new int[len]; for( int i = 0;i<len;i++){ dp[i] = gas[i]-cost[i]; sum+=dp[i]; } int num = 0; if( sum < 0) return -1; for( int i = 0;i<len;i++){ num = 0; for( int j = i;j<len;j++){ num+=dp[j]; if( num < 0){ break; } } if( num >= 0 ) return i; } return -1; } }
2、仔细看题目,发现答案其实是唯一解,那么就只需要从后向前遍历,找到最大子串和的起始位置i,(结束位置是最后一位)
在总消耗>0的前提下,返回起始位置i。
否则返回-1。
public class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int len = gas.length; int sum = 0; int max = 0; int start = 0; for( int i = len-1;i>=0;i--){ sum+=(gas[i]-cost[i]); if( sum > max ){ max = sum; start = i; } } if( sum < 0) return -1; return start; } }