leetcode14.gasStation

问题描述如下：

There are N gas stations along a circular route, where the amount of gas at stationi isgas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from stationi 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.

java代码：

package leetcode13.gasStation;import java.util.Arrays;/** * Gas Station *  * 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. */public class Solution {public static void main(String[] args) {int[] gas = {3,4,4,3};int[] cost = {4,3,2,4};System.out.println(canCompleteCircuit(gas, cost));}// O(n) DPpublic static int canCompleteCircuit(int []gas, int[] cost){int curGas=0;int total=0;int startIndex=0;for(int i=0;i<gas.length;i++){int curRemain=gas[i]-cost[i];if(curGas>=0){curGas+=curRemain;}else{curGas=curRemain;startIndex=i;}total+=curRemain;}return total>0?startIndex:-1;}}

总结：可以利用暴力的O（N^2）的方法，当然更好的办法是利用DP（动态规划）